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LABORATORY MANUAL 

i 

OF 

INORGANIC CHEMISTRY ----- 

AND 

ELEMENTARY QUALITATIVE ANALYSIS 


C. C. HEDGES 

HEAD OF DEPARTMENT OF CHEMISTRY AND CHEMICAL ENGINEERING 

H. R. BRAYTON 

PROFESSOR OF INORGANIC CHEMISTRY 
IN THE 

AGRICULTURAL AND MECHANICAL COLLEGE OF TEXAS 



D. C. HEATH AND COMPANY 

BOSTON NEW YORK CHICAGO 

ATLANTA SAN FRANCISCO DALLAS 
LONDON 


ABRIDGED TABLE 


ELEMENT 

SYMBOL 

ATOMIC 

WEIGHT 

ELEMENT 

SYMBOL 

ATOMIC 

WEIGHT 

Aluminum. 

. A1 

26.97 

Lithium.... 

. Li 

6.94 

Antimony. 

. Sb 

121.77 

Magnesium. 


24.32 

Arsenic. 


74.96 

Manganese. 

.. Mn 

54.93 

Barium. 

. Ba 

137.37 

Mercury. 

. Hg 

200.6 

Bismuth. 

. Bi 

209.00 

Molybdenum. 

. Mo 

96.0 

Boron. 

. B 

10.82 

Nickel. 

... Ni 

58.69 

Bromine. .... 

. Br 

79.92 

Nitrogen. 

. N 

14.01 

Calcium. 


40.07 

Oxygen. 

. O 

16.0 

Carbon. 

. C 

12.00 

Phosphorus. 

. P 

31.03 

Chlorine. 

. Cl 

35.46 

Platinum. 

... Pt 

195.23 

Chromium. 

. Cr 

52.00 

Potassium. 

. K 

39.10 

Cobalt. 

. Co 

58.94 

Silicon. 

. Si 

28.06 

Copper. 

. Cu 

63.57 

Silver ... 

. Ag 

107.88 

Fluorine. 

.. F 

19.00 

Sodium. 

. Na 

23.00 

Gold. 

. Au 

197.20 

Strontium. 

. Sr 

87.63 

Hydrogen. 

. H 

1.008 

Sulfur.. 

.. S 

32.06 

Iodine. 

. I 

126.93 

Tin. 

.. Sn 

118.70 

Iron. 

. Fe 

55.84 

Uranium. 

. U 

238.17 

Lead. 

. Pb 

207.20 

Zinc. 

. Zn 

65.38 


COMMON VALENCE OF ELEMENTS AND RADICALS 


i 

2 

3 

4 

5 

6 

Br 

Ba 

A1 

C 

Sb 

Cr 

Cl 

Ca 

Sb 

N 

As 

S 

Cu 

C 

As 

Pt 

N 


F 

Co 

Bi 

Si 

P 


H 

Cu 

B 

s 



I 

Fe 

Cr 

Sn 



Li 

Pb 

Au 

(Si0 4 ) 



Hg 

Mg 

Fe 




N 

Mn 

N 




K 

Hg 

Ni 




Ag 

Ni 

P 




Na 

N 

(BO,) 




(NO*) 

O 

(AsO,) 




(NO,) 

Sr 

(As0 4 ) 




(Mn0 4 ) 

S 

(PO,) 




(C*H,0*) 

Sn 

(P0 4 ) 




(OH) 

Zn 





(CIO) 

(SO,) 





(CIO*) 

(S0 4 ) 





(ClOa) 

(CO,) 





(CIO*) 

(Cr0 4 ) 

(Cr*0 7 ) 

(B 4 0 7 ) 






Note. — Student should complete this table as additional compounds are studied. 




















































LABORATORY MANUAL 

OF 

INORGANIC CHEMISTRY 

AND 

ELEMENTARY QUALITATIVE ANALYSIS 


C. C. HEDGES 

HEAD OF DEPARTMENT OF CHEMISTRY AND CHEMICAL ENGINEERING 


H. R. BRAYTON 


PROFESSOR OF INORGANIC CHEMISTRY 
IN THE 

AGRICULTURAL AND MECHANICAL COLLEGE OF TEXAS 



D. C. HEATH AND COMPANY 

BOSTON NEW YORK CHICAGO 

ATLANTA SAN FRANCISCO DALLAS 
LONDON 





Copyright, 1927, 

By D. C. Heath and Company 


©Cl A1 004428 


PRINTED IN U.S A. 


OCT-1’27 


PART I 

GENERAL INORGANIC CHEMISTRY 

AND 

ELEMENTARY QUALITATIVE ANALYSIS 








PREFACE 


This Laboratory Manual is intended for college students who devote from three to five 
hours weekly in the laboratory throughout the year to the study of Inorganic Chemistry. 
It is not claimed for the Manual that new subject material is presented. As used in the 
Agricultural and Mechanical College of Texas it serves as a guide to direct the study of 
general fundamental principles of Inorganic Chemistry. In many cases, methods of presen¬ 
tation are given which have been found most effective with our freshman classes. It is not 
the purpose of the authors to present an outline of experimental work which covers the 
entire field of chemistry. Far too many manuals are already available, from which it is 
necessary to make so many omissions in order to accommodate the time schedule, that the 
student fails to grasp the fundamentals. 

The exercises are to be detached along the perforations, and being printed on gummed 
paper, may be pasted in any type of notebook, but preferably in a permanently bound book, 
furnishing the student a permanent record of his laboratory work for reference. The in¬ 
structor may demand as much or as little written discussion of the experimental work as he 
desires, or may change the order in which the work is given and still obtain a neat continuous 
notebook. 

The sequence of experimental work fixes the order in which these topics are considered 
in the classroom and lecture. Under no conditions do we allow the practice and the theory 
to become separated. The same weekly assignments apply to each. This necessitates the 
use of some of the laboratory time to drive home principles being stressed at that particular 
time in theory. It will be noted that the order of topics differs from most outlines, in that 
symbols, valence, formulas, acids, bases, salts, nomenclature, equations, calculations, and 
weight laws are considered early in the course and considerable time devoted to these topics. 
Well grounded in these fundamentals the student is able to proceed much faster and with a 
better understanding of the work which follows. For convenience the work is outlined by 
weeks. 

The correct use of this Manual requires the “start-stop” method of instruction. When 
one or two experiments have been performed emphasizing certain properties or preparations 
or a general principle, the whole section is halted and a general discussion of results follows. 
The instructor must be on the job, full of his subject, and by explanation and questions, 
arouse the interest and enthusiasm of his students. The “cookbook” method of instruc¬ 
tion is discouraged and the student encouraged to develop an inquisitive disposition and 
reason out the principles involved in each week’s assignment. Before leaving the laboratory 
each period, the notebook is written up and the instructor checks the results. A short test 
is given each week on the preceding laboratory work. 

The experimental work is purposely made simple in order to give the student a thorough 
understanding of fundamentals. These principles are then enlarged on and developed by the 
instructor in the classroom. No complicated or expensive apparatus, which would be an 

3 


important factor with large classes, is required. In some cases, as in ionization, the instructor 
demonstrates the work to his class. 

The last third of the year is devoted to the study of the metals and at the same time the 
work in the laboratory consists of qualitative analysis. The student runs known solutions, 
separately and in groups, testing for both metals and radicals. He is then assigned unknown 
solutions. The theory of separation is discussed and the weekly tests selected from the 
reaction pages or review questions. The equations involved in the separation of the metals 
are given in the manual. The reaction pages opposite each exercise may be torn out along 
perforations and the equations balanced by the student. It is not intended that these be 
assigned for out-of-class work. After completing an unknown analysis the student turns in 
his report on the “unknown report” sheet, which is also perforated. (A number of these 
sheets will be found near the end of the Manual.) This work in elementary qualitative 
analysis is intentionally made a brief course for the purpose of stimulating interest in the 
subject, and giving the characteristic reactions of the more common metals, 

The instructor should be warned that desired results can be obtained in the use of this 
outline, only in proportion to his efforts. He must spend his full time in the laboratory, dis¬ 
cussing with his section the work to be done and the results obtained, continually checking 
the work and questioning the individual student. 

C. C. Hedges 

College Station, Texas H. R. Brayton 

March 10, 1927 


4 


GENERAL DIRECTIONS 


The purpose of laboratory work is to demonstrate, in the manner recognized as most 
scientific, the facts outlined in the text. This work, then, is not a separate course, but 
substantiates the theory. Good technique demands extreme neatness, correct manipula¬ 
tion of apparatus and rapid but thorough work. This technique can only be acquired by 
individual work and handling of apparatus. Unless otherwise stated, each student is ex¬ 
pected to perform all experiments. It is not expected that any student will consistently 
obtain absolutely correct results. It is not the function of this course to develop such a 
genius. Each student, however, should take unto himself the personal obligation to make 
all data he reports, set forth only the actual results he obtained by performance of the 
outlined work. Only by so doing will the purpose of this course be realized. 

The student will provide himself with two bound-back composition notebooks — one 
for lecture notes and one for laboratory data; a black-covered, bound-back notebook for 
permanent laboratory reports, his textbook, and this laboratory outline. Each student 
is expected to be prepared to perform the work assigned, and not spend the laboratory 
period in study. He should bring to the laboratory his textbook, laboratory outline, and 
laboratory notebooks. All questions should be answered and all data recorded. The 
notebook prepared by the student is to be an aid to his recitation work in theory and weekly 
tests on the laboratory work which are given at the beginning of each laboratory period 
on the previous work. 

The instructor will outline the work to be completed each period and during the labora¬ 
tory period will discuss with the section each experiment before and after its completion, 
together with any additions, variations, or cautions. This manual will never be a success 
with students unless each instructor teaches in the laboratory the entire practice period, 
making laboratory work the tool which drives home and clinches the theory. 


5 


INSTRUCTIONS 
WHEN ASSIGNED A DESK 

1. Secure from your instructor a tag bearing your name, section, number, and number of 
desk you have been assigned. 

2. All sections will line up numerically and present tags at stock-room window and receive 
keys. (These keys are your property for the term and must be brought to laboratory 
each period. A charge of twenty-five cents is made on keys loaned temporarily from 
the stock room.) 

3. Open desk and drawer and check apparatus with this list. If any articles are missing 
or broken, list on a pink slip secured from instructor. 

4. Instructor will 0. K. this slip and you may obtain these articles without charge at the 
stock room. (If these articles are not reported short the first laboratory period a charge 
will be made.) 

5. Begin work as directed by instructor. 

6. Any additional supplies needed which are not listed may be secured from the stock 
room by signing a blue slip at the window, listing articles, your name, initials, desk 
number, and date. 

7. If the catch which locks the drawer is out of order report same to your instructor and 
it will be repaired. AH desks and drawers must be left locked when you leave the 
laboratory. 

8. It is frequently necessary to assign two men to one desk. When this is done, the break¬ 
age bill is divided, regardless of which one signs for the apparatus. 

9. Some experiments require special apparatus which is returned the same period, such 
as thermometers, distilling flasks, condensers, weights, etc. You will sign for this 
apparatus on a white slip. When material is returned in good condition this slip will 
be given you. Any breakage of special apparatus is charged to the man signing for it. 

WHEN CHECKING IN LABORATORY DESKS 

1. Clean and dry all apparatus arranging everything in order for checking on the top of 
desk. 

2. Any apparatus broken or missing from the list must be secured from stock room, 
signing for same on blue slip. 

3. List on pink credit slip all materials in excess of that to be left in the desk, and return 
this material, clean and in good condition, together with the pink slip, to the stock room. 
Towels, litmus, matches, filter paper, glass tubing, files, and wire gauze are not returnable. 

4. Throw in waste jar all rubbish in desk. The drawer and cabinet of each desk must be 

left absolutely clean and dry. 

5. Call instructor to check desk when ready. If apparatus is incomplete or desk and 

6 


material dirty, instructor will pass to next man, and you will be forced to wait your 
turn at the end of the list. 

6. When through checking with instructor, lock drawer and desk, give key to instructor, 
wipe top of desk, arrange desk reagents, and leave laboratory in perfect condition. 

7. A charge of $1.00 is made for all keys not returned. A charge of $1.00 is made for 
cleaning apparatus and checking desk if not checked within one week after student 
has finished work. 


EACH DESK SHOULD CONTAIN THE FOLLOWING APPARATUS: 


Check 

- 4 Beakers — 100 to 250 cc. 

- 1 Clamp for Test Tubes 

- 1 Cylinder, Graduated 25 cc. 

- 2 Flasks, Erlenmeyer — 250 cc. 

-r 2 Funnels — 50 mm. 

- 1 Wire Gauze 

- 2 Glass Squares 

- 1 Mortar and Pestle 

-1 Watch Glass — 2 " 

- 1 Thistle Tube 

1 Horn Spatula Spoon 

- 1 File 

- 1 Wingtop 


Check 

-1 Bunsen Burner and Tubing 

- 1 Iron Stand and Ring 

- 1 Universal Clamp 

- 2 Mason Jars 

- 1 Rubber Stopper No. 4 

- 1 Test-tube Rack 

- 12 Test Tubes — 10 cm. 

- 1 Test-tube Brush 

-1 Wash Bottle -— 500 cc. 

- 2 Evaporating Dishes No. 00 

-1 Funnel Stand 

- 1 Crucible 


7 









1. and 2. 
3. 

4. and 5. 
6. and 7. 
8 . 
9. 
10 . 
11 . 

12. and 13. 
14. and 15. 
16. 

17. and 18. 


WEEKLY SCHEDULE OF LABORATORY WORK 

(FIRST TERM) 

Matter and Energy, Metric Units, Physical and Chemical Change. 
Oxygen. 

Symbols, Valence, Formulas, Acids, Bases, Salts, Neutralization. 
Nomenclature, Equations, Calculations. 

Hydrogen. 

Water. 

Laws of Chemical Combination, Atomic Theory, Gram-molecular-Volume. 
Three states of matter. 

Gas Laws. 

Solutions. 

Ionization. 

Sulfur and its Compounds. 


WEEKLY SCHEDULE OF LABORATORY WORK 

(SECOND TERM) 

1. Periodic Law. 

2. and 3. Atmosphere, Nitrogen, and Its Compounds. 

4. and 5. Halogen Family and Compounds. 

6. Carbon and Its Simple Compounds. 

The remainder of the second term in theory is devoted to metals. In laboratory the 
work consists of elementary qualitative analysis. 


8 


LABORATORY OUTLINE OF INORGANIC CHEMISTRY 



INTRODUCTION 

(FIRST WEEK) 

1. Operation of the Bunsen Burner. — Unscrew the chimney and examine the construction of 
the burner. Attach burner base to gas tap by means of rubber tubing. Open gas inlet by means 
of screw controlling the needle valve. Light the gas and slowly close gas inlet screw. W hat is the 
effect on the flame (1)? Attach the chimney to the burner, open the gas inlet screw, and screw 
the chimney as far down as possible. Light the gas and slowly unscrew the chimney. What is the 
effect on the flame (2)? What is the cause of a flame “ striking back ” (3)? By means of a copper 
or iron wire determine which flame is hotter — the luminous or non-luminous (4). Llold a porce¬ 
lain dish in both luminous and non-luminous flame. Explain the deposit in the one case (5). 


METRIC UNITS 



(See table in Appendix) 

2. Length. — Express dimensions in millimeters, centimeters, and meters of the following: 


Article 

mm. 

cm. 

m. 

T^st HTuVip, . 

• 



TTpaker . 

Funnel . 

Wht p h Class . 

M ason .Tar . 



Express in meters and kilometers the following: 



m. 

km. 


Width 0 f Laboratory . 




1 HD Vfirds . 

Feet . 

440 Varrls . 

2 Miles . 



9 





















































3. Volume. Measuring the water in your graduated cylinder, record as cubic centimeters 
and liters the volume of the following pieces of apparatus filled to ordinary carrying capacity: 


Article 

cc. 

1 . 


Test Tube. 




Beaker. 

Erlenmeyer Flask. 

Wash Bottle. 

Mason Jar. 


Pour into your beaker the amount of water you estimate to be 2 cc., 10 cc., 25 cc., 100 cc. After 
each estimate measure the amount you have used to determine whether you have an approximate 
idea as to these volumes. 


4. Weight. — Weigh on clean filter papers on the pan of the balance, two grams'of each of the 
following substances: sodium chloride, sulfur, pulverized iron, wood charcoal, and barium 
chloride. Arrange these according to their comparative bulk and record this order from lightest 
to heaviest (1). Return these substances to the proper shelf bottle. 

Record your weight in pounds and kilograms (2). An athlete throws a 16-pound shot. Express 
this in grams (3). 


5. Density of Water. Using the laboratory balance make the following weighings and 
record all data: 


Vol. of Water 

Total Weight 

Weight of Water 

Weight of 1 cc. 
of Water 

Erlenmeyer flask (empty) 

Erlenmeyer flask + 10 cc. 

Erlenmeyer flask + 25 cc. 

Erlenmeyer flask + 50 cc. 





10 



























































































6. Density of Solids. — If we use water as the standard of density of solids and liquids, it is 
only necessary to know the volume and weight of any substance to calculate its absolute density 
which is the weight in grams of one cubic centimeter of the substance. If the substance is regular 
in shape the volume can be calculated from dimensions; if of irregular shape, the volume can be 
determined by immersion in water, observing the volume of water displaced. 

Using tap water, a graduated cylinder, laboratory balance, and meter stick, record necessary 
data and calculate density of: 


Metal used 

Weight 
in gms. 

Water 
used in cc. 

Water in 
cc. after 
immersion 

Dimension 
in cm. 

Volume 
in cc. 

Density 

Mossy zinc. 

Lead strip. 

Unknown cube. 

Unknown cylinder. 








7. Density of Liquids. — By carefully measuring a definite volume of a liquid in cubic centi¬ 
meters, and weighing this, the density of the liquid (weight per cubic centimeter) may be cal¬ 
culated. 

If 8 cc. of mercury weighs 108.8 gms., calculate its density (1). If the density of mercury is 
13.6, calculate the weight of 200 cc. of the metal (2). If the density of mercury is 13.6, calculate in 
cubic centimeters the volume of 95.2 grams of the metal. 


8. Density of Gases. — The absolute density of a gas may be calculated by comparing the 
weight of one cubic centimeter of the gas (obtained as with liquids) with the weight of one cubic 
centimeter of water. However, since the weight of a cubic centimeter of a gas will be such a small 
figure, we frequently compare the weight of a gas with the weight of an equal volume of some other 
gas taken as standard instead of water. This comparative weight is called its relative density. 

The weight of one liter of oxygen is 1.429 gms. and of one liter of air is 1.2928 gms. Calculate 
the absolute density of oxygen and of air (1). Calculate the relative density of oxygen compared 
with air (2). 


11 
























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PHYSICAL AND CHEMICAL CHANGE 


(SECOND WEEK) 

The final proof as to whether or not a chemical change has occurred is whether or not a new 
substance has been formed. The chemical and physical properties of the substances before and 
after the action tell us whether we have the same or new substances. 

Common physical tests we use are color, odor, taste (caution), density, hardness, brittleness, 
solubility, conductivity, etc. Chemical tests employed are those which have been found to be 
characteristic for each substance, and therefore must be varied. Characteristic tests are the action 
of various reagents, identification of gases liberated, residue left, or precipitate thrown down. 


9 . Determine by physical and chemical tests whether chemical reaction occurs in the following: 

(a) Drop a splinter of wood in a test tube containing a little dilute hydrochloric acid. Does 
there appear to be any change in the wood or the acid (1)? Burn a small piece of wood 
in a porcelain crucible until a gray ash is obtained. When cool, add a little dilute hy¬ 
drochloric acid, as before. Does there appear to be any change in the ash or the acid 
(2)? Is a gas liberated (3)? Comparing the behavior of wood and the wood ash, using 
the same reagent, would you conclude that the burning of wood is a chemical or a 
physical change (3)? 

(b) Repeat (a) using a small piece of magnesium ribbon in place of the wood. Answer the 
same questions (1), (2), (3). 

(c) Burn a small pinch of sulfur in a porcelain dish. Note the odor. Is a residue left (1)? 
This characteristic brimstone odor when sulfur is burned is a chemical test for sulfur. 
Has the sulfur changed chemically (2)? 

(d) To a very small pinch of sulfur in a test tube add enough carbon bisulfide (very 
inflammable) to dissolve. Pour a few drops of the clear liquid on a watch glass and 
allow to evaporate in air (do not heat). Is there a residue (1)? Has the sulfur 
changed chemically (2)? Is sulfur soluble in water (3)? 

(e) Dissolve a pinch of sugar in water. Can the sugar be recovered as was sulfur in (d) 
(1)? Is sugar soluble in carbon bisulfide (2)? Heat a small amount of dry sugar in a 
test tube. Note any change. Is there a residue? If so, test its solubility in water (3). 
Is a gas driven off (4)? Has a chemical change taken place (5)? 

(f) In the following state whether or not a chemical change occurs: Water is boiled (1); 
paper is burned (2); ice is melted (3); dynamite is exploded (4); iron is rusted (5); 
wood is decayed (6); bread dough is baked (7); air is breathed (8); an electric globe 
is switched on (9); a radiolite watch glows in the dark (10). 


GLASS WORKING 

10 . Obtain from the stock room two feet of glass tubing. Moisten edge of file, lay glass tubing 
on desk, and make one sharp scratch at the point where you wish to cut the tubing. Hold in both 
hands and apply pressure by the thumbs from beneath the scratch, at the same time each hand 
exerts a slight pulling force from each side of the scratch. Round all edges of freshly cut tubing 
by heating the ends in the non-luminous flame of your burner until the rough edges are fused 
together. 


12 








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. 

. 

. 


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11 . Bend a short piece of glass tubing by holding lengthwise in the flat luminous flame of the 
wingtop burner. Rotate the tubing slowly between the fingers while heating, so that all portions 
are at the same temperature. When the weight of the end causes the tubing to begin to bend, 
remove from flame and carefully bend to right angle, keeping both ends in the same plane. For 
the sake of comparison try heating and bending a piece of glass tubing without using the wingtop 
on your burner. 


12 . Heat about six inches of glass tubing in the middle with an ordinary burner flame, rotate 
between the fingers, and when the glass begins to bend, remove from flame and pull the two ends 
apart. This gives you a fine capillary tube which may be cut with a file leaving a small noazle 
which will be used later. Save all pieces of bent tubing. 


13 










. fj 





OXYGEN 


(THIRD WEEK) 

13 . (Two Students.) In a large hard glass test tube place about 15-20 gms. of a mixture of 
ferric oxide and potassium chlorate. Run a delivery tube through a one-hole stopper and fasten 
test tube in clamp on your iron stand. Fill a pneumatic trough with water. Heat test tube gently 
and collect the gas liberated in bottles which have been filled with water and inverted in the 
trough. Collect four bottles of the gas, cover with glass squares, and set aside for (14). When 
discontinuing heating of the tube remove end of delivery tube from the water. Why (1)? 


14 . (Two Students) 

(a) Insert a glowing splinter in one bottle of the gas (1). 

(b) Into second bottle lower a combustion spoon (stock room) containing a pinch of 
sulfur which has been ignited in a flame. What are the fumes (1)? After combustion 
put a few drops of water in bottle containing these sulfurous fumes, close top with 
hand, shake, and then test water with a piece of blue litmus paper. Explain (2). 

(c) Repeat (b) substituting a small pinch of red phosphorus for the sulfur. What are 
these fumes (1)? What is the effect on litmus when these fumes are absorbed in a 
fit tie water (2)? Explain (3). 

(d) Heat a small piece of picture wire, the end of which has been slightly unraveled. (If 
wire is heated and dipped in a little sulfur and the sulfur ignited, the wire will hold 
heat.) Hold the glowing wire in the fourth bottle of oxygen. What becomes of the 
iron and oxygen (1)? 


15 . In test tubes heat separately one gram of each of the following substances and test the 
gas liberated by means of a glowing splinter: Barium peroxide, mercuric oxide, potassium nitrate, 
potassium chlorate, manganese dioxide, ferric oxide, lead peroxide, and a mixture of potassium 
chlorate and manganese dioxide. In heating substances in a test tube, do not point the mouth of 
the tube toward your face or toward your neighbor. Use test-tube clamp in holding tube. Which 
of these substances liberate oxygen (1)? 


16 . Using your test-tube holder, ignite the tip of a piece of magnesium ribbon in a flame. Is a 
residue left (1)? If so, has a chemical change taken place, and what has been formed (2)? Why 
will some substances burn in a bottle of oxygen, and not burn in air as does the magnesium (3)? 


17 . Secure from the stock room in an evaporating dish, a very small piece of yellow phosphorus. 
(Caution: — Do not handle phosphorus with the fingers because it will cause a painful burn.) 
Dissolve in a little carbon bisulfide and pour a little of this solution on a dry paper which has been 
placed on a glass square. Allow the carbon bisulfide to evaporate in air and note the effect (1). 
Do not handle the paper with the fingers. When through, burn all pieces of phosphorus and wash 
any remaining solution down the sink. Wdiat would happen if a towel were used in mopping up 
phosphorus and phosphorus solution (2)? 


14 



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Vi. c <../.<• i>;u ,i v . 4 




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■ 

ts<d ■ . ■ . ■ * vr 



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18 . What type of compound is formed when metals are burned in air or oxygen (1)? When 
nonmetals are burned in air or oxygen (2)? What is an oxide (3)? What is oxidation (4)? 
What is combustion (5)? Distinguish between ordinary combustion and spontaneous com¬ 
bustion (6). Give two illustrations of each in everyday life (7). 


19 . Is oxygen lighter or heavier than air (1)? How could this be demonstrated (2)? Does 
oxygen burn (3)? Does it support combustion (4)? List four uses of oxygen in daily life (5). 


Note: As tjie student progresses with his work he should cultivate his power of observation, 
develop an inquisitive disposition, and train himself to act in emergencies. For example: 

Give names and formulas of your desk reagents. Are these solutions dilute or concentrated? 
What is the system employed in arrangement of shelf reagent bottles? Are the liquids dilute or 
concentrated? How and where are the desk bottles filled? What is the source of our laboratory 
gas? What is the correct method of pouring a liquid from a stoppered bottle? How is a filter 
paper folded? How do you force glass tubing through a hole in a rubber stopper? Where is the 
nearest fire extinguisher? How would you extinguish a phosphorus fire? Where is the first-aid 
cabinet? How treat the following: Acid in the eye, alkali in the eye, an acid burn, an alkali 
burn, a flame burn, acid on clothes, alkali on clothes, acid swallowed, alkali swallowed, a fainting 

spell, cuts? , 

If unable to answer these various questions, do not hesitate to ask your instructor for assistance, 
for the occasion may unexpectedly arise when the sight, or even the life of your fellow worker, may 
be saved by quick action. 


15 






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SYMBOLS, VALENCE, FORMULAS, ACIDS, BASES, SALTS, NEUTRALIZATION, 
EQUATIONS, NOMENCLATURE, CALCULATIONS 

(FOURTH, FIFTH, SIXTH, AND SEVENTH WEEKS) 

Note: Extra laboratory time will be devoted to theory. 

Summary. In the early part of our text we learn that all matter may be classified into ele¬ 
ments, compounds, and mixtures. These may exist in the solid, liquid, or gaseous form, the three 
states of matter, depending on the nature of the substance, temperature, and pressure. Ele¬ 
ments are further classified into metals, nonmetals, and metalloids. Depending on their chemical 
and physical properties we say they are active or inactive. When two or more elements combine 
chemically to form a new substance we call this a compound. 

Chemical Reactions. When new substances are formed the process by which the change is 
accomplished is a chemical reaction. If we cause elements to combine to form a compound which is 
more complex than the elements themselves, this type of chemical reaction is spoken of as direct 
union or composition or synthesis. Under certain conditions — application of heat, use of a 
catalytic agent, or an electric current, etc., compounds may be broken down into simpler com¬ 
pounds or elements. Such a chemical change is called decomposition or analysis. In many cases 
one simple substance reacts with one compound substance to form a new simple substance and a 
new compound substance. Such a chemical change is called displacement. W hen two compounds 
react on each other to form two new compounds the chemical change is called double decomposi¬ 
tion. 

In naming chemical substances there has been no definite system or plan used. Many elements 
have been named from the Greek or Latin word illustrating a property or occurrence of the ele¬ 
ment. For example: oxygen (acid former), hydrogen (water former), helium (the sun), chlorine 
(yellowish-green), etc. As a rule metals end in -um or -ium as potassium, sodium, ferrum (iron), 
argentum (silver), etc. Nonmetals usually end in -ine, -gen, or -on, as nitrogen, iodine, silicon. 

As a means of simplifying our work in chemistry we use abbreviations for elements and com¬ 
pounds as a shorthand method of expression. The abbreviation for an element is called a symbol. 
The abbreviation for a compound is called a formula. A collection of symbols or formulas ex¬ 
pressing a chemical change is called an equation. 

Symbols for elements are usually the first letter of the word, the first and second letter, or the 
first and third letter: P (phosphorus), N (nitrogen), Ca (calcium), Mg (magnesium), Mn (man¬ 
ganese), etc. In many cases the letters are taken from the Latin, Greek, Spanish, or German 
words: Fe (ferrum), Au (aurum), Pt (plata), Hg (hydrargyrum), Sb (stibium), Na (natrium), 
K(kalium), etc. It is important that we learn the symbols of the common elements early in the 
course. Your instructor will mention the elements with which you should become familiar. At 
the same time you should know the exact meaning of the symbol. Thus H stands for one atom of 
hydrogen or 1.008 parts by weight; Fe stands for one atom of iron or 55.84 parts by weight; 2Na 
stands for two atoms of sodium or 46 parts by weight. Since the atomic weights given in our table 
are only comparative weights of the different elements, we may assume these figures to be- grams 
or ounces or pounds or tons and solve any problem directly without conversion. In our work we 
use the gram as our metric unit of weight, and the atomic weight of the element is then the gram- 
atomic-weight. 

Formulas of compounds are groups of symbols which when put together tell you (1) the ele¬ 
ments present in the compound, (2) the number of atoms of each element present, (3) the weight 
of each element present. Thus H 2 SO 4 tells us the three elements hydrogen, sulfur, and oxygen 
are present in the compound; that there are two atoms of hydrogen, one atom of sulfur, and four 
atoms of oxygen; and referring to our table of atomic weights we find that the formula tells us 
there are 2 X 1.008 grams of hydrogen, 32 grams of sulfur, and 4 X 16 grams of oxygen. Therefore 
the gram-molecular-weight of the compound H 2 S0 4 is 98.016. 


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Simple calculation will tell us the percentage composition of the compound. The percentage 
of the various elements in H 2 S0 4 will be (2.016 -5- 98.016) X 100% = Hydrogen, (32 -4- 98.016) 
X 100% = Sulfur, and (64 -5- 98.016) X 100% = Oxygen. 

The reverse calculation is also simple. If a chemical analysis tells us the percentage of each 
element in a compound, we can refer to our table of atomic weights and calculate the number of 
atoms of each element and write the simple formula. For example: Chemical analysis shows a 
certain compound contained 11.32% of carbon, 45.29% of oxygen, and 43.39% of sodium. Since 
percentage by weight means grams of substance in 100 grams of the compound, this compound 
will contain in every 100 grams 11.32 gms. of carbon, 45.29 gms. of oxygen, and 43.39 gms. of 
sodium. Using our table of atomic weights to determine the number of atoms of each element in 
this compound, we find that 43.39 - 4 - 23 = 1.88 atoms of sodium, 11.32 - 4 - 12 = .94 atoms of 
carbon, and 45.29 - 4 - 16 = 2.82 atoms of oxygen. Our formula then is Na^C^O,^. Chemical 
theory, however, tells us there are no fractions of atoms, so expressing this in whole number of 
atoms, dividing through by .94 we derive the formula Na 2 C0 3 . 

In interpreting a formula, the small subnumber multiplies only the element or group which 
it follows. A large number placed before a compound multiplies every element in the compound. 
Thus H3PO4 means three atoms of hydrogen, one atom of phosphorus, and four atoms of oxygen, 
while 2 H 3 PO 4 doubles the amount of each element. 


ACIDS, BASES, SALTS 

20 . You have prepared some oxides of nonmetals which are soluble in water, giving a solution 
which affects litmus paper. List three such oxides (1). Are these chemical reactions (2)? What 
type of compound is formed (3)? 


21 . Test separately a little of each of the following with red and blue litmus: Hydrochloric 
acid, sulfuric acid, phosphoric acid, hydrobromic acid, nitric acid, hydriodic acid and vinegar. 
What is the effect on litmus (1),? Dilute 1 cc. of hydrochloric acid with 10 cc. of water. Dip a 
clean glass rod in this solution and taste it, rinsing the mouth with water immediately. All dilute 
acids have this taste and similar effect on litmus. From the formulas on the bottles, what ele¬ 
ment is present in all acids ( 2 )? 


22 . When very active metals such as potassium, sodium, calcium, and magnesium are burned, 
the respective oxides are formed. These oxides are soluble in water. Less active metals such as 
iron and copper when burned in oxygen produce oxides which are insoluble in water. Drop a pinch 
of each of the following in separate test tubes half full of water: Sodium oxide, calcium oxide, 
copper oxide, iron oxide. W T arm and filter into clean test tubes and test the filtered solutions with 
both red and blue litmus: What is the effect (1)? Have chemical reactions taken place (2)? What 
type of compound is produced (3)? 


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23 . Test separately a little of each of the following solutions with both red and blue litmus: 
Sodium hydroxide, potassium hydroxide, calcium hydroxide, and ammonium hydroxide. What 
is the effect on litmus (1)? 

Dilute 1 cc. of sodium hydroxide with 10 cc. of water. Dip a clean glass rod in this solution 
and taste it, rinsing the mouth with water immediately. All dilute bases have a similar taste 
and the same effect on litmus. From the formula on the bottles, what elements are present in all 
bases (2)? 


24 . Since bases and acids act oppositely on litmus it is possible to determine when a solution 
is neutral, containing neither acid nor base in excess. Instead of litmus paper it is often more con¬ 
venient to use the indicator in liquid form. Two indicators commonly used are phenolphthalein 
and methyl orange. Try the effect of a few drops of each of these indicators on dilute solutions of 
both acids and bases. Record color changes (1). Since replaceable hydrogen is the acid constitu¬ 
ent of all acids and hydroxide is the basic constituent of all bases, the indicator test shows whether 
hydrogen or hydroxide groups are present. Whenever possible in a solution the hydrogen and 
hydroxide groups will combine to form HOH, commonly written H 2 O and called water. Pure 
water has no effect on indicators. The process of combining an acid with a base, causing the 
hydrogen of the acid and the hydroxide of the base to unite to form water, is called neutralization. 

In addition to the hydrogen we have present in every acid a nonmetal or group of nonmetals 
which we call the acid radical. Every base is made up of a metal combined with the hydroxide 
group. When neutralization takes place and the hydrogen and hydroxide groups combine, the 
rest of the acid and the base combine to form a salt. A salt, then, contains the metal of the base 
combined with the acid radical of the acid. 

In a clean evaporating dish place 5 cc. of your desk hydrochloric acid and two drops of phe¬ 
nolphthalein. Slowly add sodium hydroxide solution drop by drop, stirring the solution in the 
Hish with a glass rod. When the first faint pink remains permanently in the dish, the acid has 
been neutralized. What two substances are formed (2)? Evaporate this solution to dryness. 
What is the residue (3)? Note taste (4). 

Test solutions of the following salts by an indicator: Calcium chloride, sodium sulfate, 
sodium nitrate, potassium chloride. What is indicated (5)? As a summary give definition of acid, 
base, and salt (6). Give tests for an acid and a base (7). 


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VALENCE AND BALANCING EQUATIONS 


In writing formulas of compounds by combining two or more elements you will find that in 
many cases two or three or even more atoms of one element will combine with a single atom of 
another element. In writing formulas of salts you will notice that the hydrogen of an acid is 
replaced by a metal. There is always a definite system of interchange or trading of one element for 
another just as in daily life we exchange a dollar bill or silver dollar for two halves, four quarters, 
one half and two quarters, etc. Just as in daily life we must have a standard for money values, in 
the same way we must have a standard of exchange or replacement for different elements. This 
trading value of different elements is called their Valence. Since hydrogen is found in all acids, 
bases, and in water, all of which are very common compounds, we may select hydrogen as a stand¬ 
ard and assume its valence as 1. From the compound H 2 0, in which two atoms of hydrogen com¬ 
bine with one atom of oxygen, we would then say that oxygen has a valence of 2. When magnesium 
was burned and the compound magnesium oxide (MgO) formed, since oxygen has a valence of 2, 
magnesium must also have a valence of 2. In the compound MgS0 4 if magnesium has a valence 
of 2, the acid radical (S0 4 ) must also have a valence of 2. We may check this from the 
formula of sulfuric acid, H 2 S0 4 , where two atoms of hydrogen each with a valence of 1 com¬ 
bine with (S0 4 ). Always represent valence by Roman numerals above the element or radical. 

For drill write the valence of the hydrogen and radicals in Roman numerals in the following 
acids: HC1, HBr, HI, H 2 S0 4 , H 2 S0 3 , HN0 3 , HN0 2 , HMn0 4 , H 2 Cr0 4 , H 3 B0 3 , H 3 P0 4 , H 3 P0 3 , 
HsAsO-s, HsAs0 4 , H 4 Si0 4 . 

We found in discussing neutralization that hydrogen and hydroxide group combined to form 
H(OH). Therefore if hydrogen has a valence of 1, the hydroxide group must also have a valence 
of 1. For further drill write the valence in Roman numerals above the metal and the hydroxide 
group in the following bases and hydroxide compounds: NaOH, KOH, LiOH, Ca(OH) 2 , Al(OH) 3 , 
Mg(OH) 2 , Ba(OH) 2 , Fe(OH) 2 , Fe(OH) 3 , Zn(OH) 2 , Sb(OH) 3 , Bi(OH) 3 , Mn(OH) 2 , NH 4 OH, 
Cr(OH) 3 . 

t For further drill, remembering that the hydrogen of an acid will combine with the (OH) group 
of a base, write chemical equations showing the action when each of the fifteen acids above react 
on each of the fifteen hydroxides above; for example: HC1 + NaOH, HC1 + KOH, HC1 + LiOH 
etc., making a total of 225 equations. In each case before completing and balancing the equation 
place the valence above the hydrogen, the acid radical, the metal, and the hydroxide group. When 
the hydrogen and hydroxide groups are equal in number and the trading values of the hydrogen 
and metal are the same the equation is balanced. 

In balancing the equations above, since the formulas given are correct, if more hydrogen or 
hydroxide are needed, the entire compound must be multiplied by some number which will give 
the required amount. 

The reactions above are of the general type reaction "acid plus base give salt plus water.” 
You have found that metals when burned in oxygen always give metallic oxides. For further drill 
write the formulas of the fifteen oxides formed when the metals of the hydroxides listed above com¬ 
bine with oxygen, using the same valence of the metal as shown in the base. Oxygen has a con¬ 
stant valence of 2. 


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SUMMARY OF PROCEDURE TO FOLLOW IN BALANCING EQUATIONS 

1. Decide to what general class each reacting substance belongs. 

2. Decide what general chemical reaction is involved. 

3. Write skeleton equation showing what elements and radicals are combined. 

4. Place Roman numerals above each element and radical on the left and transfer same valences 
to the right. 

5. Write true formulas of each substance satisfying all valences. 

6. Lock each true formula by encircling with a ring. 

7. Balance the atoms on left and right by multiplying outside the ring. 

8. Check the number of atoms on both sides. 


SOME GENERAL CHEMICAL REACTIONS 

1. Metal + oxygen (heated) = metallic oxide. 

2. Nonmetal + oxygen (heated) = nonmetallic oxide. 

3. Many metallic oxides (heated) = metal + oxygen. 

4. Peroxides heated = oxides + oxygen. 

5. Soluble metallic oxide (basic anhydride) + H 2 0 = base. 

6. Soluble nonmetallic oxides (acid anhydride) + H 2 0 = acid. 

7. Metallic oxides + reducing agent = metal + new oxide. 

8. Salts rich in oxygen + reducing agent = new salt -f oxide. 

9. Many salts + oxidizing agent = new salts richer in oxygen. 

10. Very active metal + H 2 0 = base + hydrogen. 

11. Very active nonmetal + H 2 0 = acid + oxygen. 

12. Active metal + acid = salt + hydrogen. 

13. Zinc or aluminum + strong base = salt + hydrogen. 

14. Active metal + active nonmetal = salt. 

15. Salts rich in oxygen heated = new salt + oxygen. 

16. Acid + base = salt + H 2 0. 

17. Acidic oxide + basic oxide = salt. 

18. Salt -f- non-volatile acid = new salt + new acid. 

19. Salt + salt = two new salts. 

20. Metallic oxide + acid = salt + H 2 0. 

21. Nonmetallic oxide + base = salt + H 2 0. 

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DRILL IN VALENCE AND BALANCING EQUATIONS 


Complete equations, balance, and mark valence above each element and radical in Roman 
Numerals. 


1 . Cu + 0 2 = 

48. Na + H 2 S0 4 — 

95. 

Mg(OH ) 2 + HN0 3 = 

2. Fe + O 2 = 

49. Ca + H 3 P0 4 = 

96. 

Al(OH ) 3 + HC1 = 

3. Mg + 02 = 

50. Mg + HC1 = 

97. 

Al(OH ) 3 + h 2 so 4 = 

4. Zn + O 2 = 

51. Zn + H 2 S0 4 = 

98. 

Al(OH ) 3 + H 3 P0 4 = 

5. A1 + O 2 = 

52. A1 + H 2 S0 4 = 

99. 

Al(OH ) 3 + HNO 3 = 

6 . Hg + O 2 = 

53. Fe + HC1 = 

100 . 

k 2 o + so 3 = 

7. Ca + O 2 = 

54. K + H 2 S0 4 = 

101 . 

K 2 0 + co 2 = 

8 . Na + O 2 = 

55. Na + HC1 = 

102 . 

k 2 o + so 2 = 

9. Ba + O 2 = 

56. Ca + HC1 = 

103. 

K 2 0 + P 2 O 5 = 

10 . K + O 2 = 

57. Mg + H 2 S0 4 = 

104. 

K 2 0 + As 2 0 3 = 

11. S + O 2 = 

58. Zn + HC1 = 

105. 

Na 2 0 + S0 3 = 

12. C + O 2 = 

59. A1 + HC1 = 

106. 

Na 2 0 + C0 2 = 

13. P + 0 2 = 

60. Fe + H 2 S0 4 = 

107. 

Na 2 0 + S0 2 = 

14. N + O 2 = 

61. K + H 3 P0 4 = 

108. 

Na 2 0 + P 2 Os = 

15. As + 0 2 = 

62. Na + H 3 P0 4 = 

109. 

MgO + S0 3 = 

16. HgO + heat = 

63. Ca + H 2 S0 4 — 

110 . 

MgO + C0 2 = 

17. Ag 2 0 + heat = 

64. Mg + H 3 P0 4 = 

111 . 

MgO + S0 2 = 

18. Ba0 2 + heat = 

65. Zn + R 3 P0 4 = 

112 . 

MgO + P 2 0 5 = 

19. H 2 0 2 + heat — 

66. A1 + H 3 P0 4 = 

113. 

CaO + S0 3 = 

20 . Na 2 0 2 + heat = 

67. Fe + H 3 P0 4 = 

114. 

CaO + C0 2 = 

21. Pb 3 0 4 + heat = 

68. Zn + NaOH = 

115. 

CaO + S0 2 = 

22. Pb0 2 + heat = 

69. A1 + NaOH = 

116. 

CaO + P 2 0 5 = 

23. K 2 0 2 + heat = 

70. Zn + KOH = 

117. 

CaO + As 2 0 3 = 

24. 0 3 + heat = 

71. A1 + KOH = 

118. 

A1 2 0 3 + S0 3 = 

25. K 2 0 + H 2 0 = 

72. Na + Cl 2 = 

119. 

A1 2 0 3 + CO 2 = 

26. Na 2 0 + H 2 0 = 

73. K + I 2 = 

120 . 

A1 2 0 3 + SO 2 = 

27. ZnO + H 2 0 = 

74. Mg + Br 2 = 

121 . 

A1 2 0 3 + P 2 O 5 = 

28. CaO + H 2 0 = 

75. Ca + F 2 = 

122 . 

NaN0 3 + H 2 S0 4 = 

29. MgO + H 2 0 = 

76. Na + I 2 = 

123. 

KN0 3 + h 2 so 4 = 

30. S0 2 + H 2 O = 

77. Mg + Cl 2 = 

124. 

MgCl 2 + H 2 S0 4 = 

31. SO 3 + H 2 O = 

78. KCIO 3 + heat = 

125. 

FeCl 3 + H 2 S0 4 = 

32. C0 2 + H 2 0 = 

79. KN0 3 + heat = 

126. 

CaC0 3 + H 2 S0 4 = 

33. P 2 O 3 + H 2 0 = 

80. NaN0 3 + heat = 

127. 

NaN0 3 + H 3 P0 4 = 

34. P 2 0 5 + H 2 0 = 

81. NaOH + HC1 = 

128. 

kno 3 + h 3 po 4 = 

35. As 2 0 3 + H 2 0 = 

82. NaOH + H 2 S0 4 = 

129. 

MgCl 2 + H 3 P0 4 = 

36. K + H 2 0 = 

83. NaOH + H 3 P0 4 = 

130. 

FeCl 3 + H 3 P0 4 = 

37. Na + H 2 0 = 

84. NaOH + HN0 3 = 

131. 

CaC0 3 + H 3 P0 4 = 

38. Ca + H 2 0 = 

85. KOH + HC1 = 

132. 

NaN0 3 + KC1 = 

39. Mg + IPO = 

86. KOH + H 2 S0 4 = 

133. 

NaCl + KN0 3 = 

40. Zn + H 2 0 = 

87. KOH + H 3 P0 4 = 

134. 

MgCU + FeS0 4 = 

41. A1 + H 2 0 = 

88. KOH + HN0 3 = 

135. 

A1 2 (S0 4 ) 3 + NaCl = 

42. Fe + steam = 

89. Ca(OH ) 2 + HC1 = 

136. 

K 2 0 + HC1 = 

43. Cl 2 + H 2 0 = 

90. Ca(OH ) 2 + H 2 S0 4 = 

137. 

K 2 0 + hno 3 = 

44. Br 2 + H 2 0 = 

91. Ca(OH ) 2 + H 3 P0 4 = 

138. 

k 2 o + h 2 so 4 = 

45. I 2 + H 2 0 = 

92. Ca(OH ) 2 + HN0 3 = 

139. 

k 2 o + h 3 po 4 = 

46. F 2 + H 2 0 = 

93. Mg(OH ) 2 + HC1 = 

140. 

Na 2 0 + H 3 P0 4 = 

47. K + HC1 = 

94. Mg(OH) 2 + H 2 S0 4 = 

141. 

Na 2 0 + H 2 S0 4 = 


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142. Na 2 0 + HN0 3 = 

143. Na 2 0 + HC1 = 

144. CaO + HC1 = 

145. CaO 4“ H 2 S0 4 = 

146. CaO -f- H 3 PO 4 = 

147. CaO -\*' HNO 3 = 

148. MgO 4- HC1 = 

149. MgO 4- H 2 S0 4 = 

150. MgO 4- H 3 P0 4 = 

151. MgO 4- HNO 3 = 

152. A1 2 0 3 + HC1 = 

153. A1 2 Os 4- H 2 S0 4 = 

154. Al 2 Os + H 3 P0 4 = 

155. AI 2 O 3 4- HNO 3 = 

156. S0 2 4- KOH = 

157. S0 2 4- NaOH = 

158. S0 2 4- Ca(OH ) 2 = 

159. S0 3 + NaOH = 

160. S0 3 + KOH = 

161. S0 3 + Ca(OH ) 2 = 


162. C0 2 + NaOH = 

163. C0 2 4- KOH = 

164. C0 2 4- Ca(OH) 2 = 

165. P 2 0 6 + NaOH = 

166. P 2 0 5 4- KOH = 

167. P 2 0 6 4- Ca(OH) 2 = 

168. P 2 0 3 4- NaOH = 

169. P 2 0 3 + KOH = 
170 v P 2 0 3 4- Ca(OH) 2 = 

171. As 2 0 3 4- NaOH = 

172. As 2 0 3 4- KOH = 

173. As 2 0 3 + Ca(OH) 2 = 

174. S0 3 4- Al(OH) 3 = 

175. P 2 0 5 4- Al(OH) 3 = 

176. C0 2 4- Al(OH) 3 = 

177. CuO 4- H 2 = 

178. CuO + C = 

179. MgC0 3 4- HC1 = 

180. CaC0 3 4- H 3 P0 4 = 

181. Na 3 P0 4 4- H 2 S0 4 = 


182. Pb(N0 3 ) 2 4- H 2 S = 

183. CdS0 4 + H 2 S = 

184. Ag 2 S 4- HNO3 = 

185. FeCl 3 4- NH 4 OH = 

186. MgS0 4 4- H 3 P0 4 = 

187. Ba0 2 4~ H 2 S0 4 = 

188. HgN0 3 4- HC1 = 

189. CuS0 3 4- H 2 S = 

190. BiCls + NaOH = 

191. Sb(N 0 3 ) 3 4- KOH 

192. HgNOs 4- KI = 

193. HgO 4~ Zn = 

194. Sb 2 0 3 4- A1 = 

195. Cr 2 0 3 + A1 = 

196. FeS0 4 + 0 2 = 

197. KCIO3 4- C = 

198. KNO3 4- C = 

199. KCIO + 0 2 = 

200. KC10 2 4- 0 2 = 


22 








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NAMING OF COMPOUNDS 


From the work just completed — writing of formulas of acids, bases, and salts — calculation 
of percentage composition and formula, and balancing of equations, you have perhaps noticed 
that once you have a working knowledge of valence, all this is possible without a knowledge of 
the names of the various substances used and formed. As our theory work is developed, however, 
it becomes necessary to use chemical terms or language to simplify our discussion. These terms 
will be taken up as we come to them in theory. We may save ourselves some worry and future 
trouble, however, if we begin to familiarize ourselves with a few common prefixes and suffixes 
which are used in our work. 

(1) A compound made up of just two elements ends in the suffix “ide” which is attached to the 
second element mentioned. Thus: (MgO) magnesium oxide, (Na 2 0) sodium oxide, (NaCl) sodium 
chloride, (HC1) hydrogen chloride, (ZnS) zinc sulfide, (NH 3 ) nitrogen hydride, (Ca 3 P 2 ) calcium 
phosphide, etc. 


(2) Where elements have more than one valence under different conditions, the suffix “ic” 
is used for the higher valence and “ous” for the lower valence. Thus FeCl 3 is ferric chloride and 
FeCl 2 is ferrous chloride; CuO is cupric oxide and Cu 2 0 is cuprous oxide; H 2 S 04 is sulfuric acid 
and H 2 S0 3 is sulfurous acid, etc. 

(3) Where more than the normal amount of an element is present than is required to satisfy 
the common valence, the prefix “per” is attached to the element affected. Thus H 2 0 is hydrogen 
oxide while H 2 0 2 is hydrogen peroxide; BaO is barium oxide and Ba0 2 is barium peroxide. 

(4) Acids of just two elements have the prefix “hydro ” and suffix “ic.” Thus (HC1) hydro¬ 
chloric acid, (HBr) hydrobromic acid, (HI) hydriodic acid, (H 2 S) hydrosulfuric acid. All such acids 
form salts ending in “ide.” 

(5) Acids containing oxygen which end in “ic” will form salts which end in “ate,” when the 
hydrogen is replaced by a metal. Thus (H 2 S0 4 ) sulfuric acid will always form sulfate salts which 
always contain the acid radical (S0 4 ); thus (Na 2 S0 4 ) sodium sulfate, (CaS0 4 ) calcium sulfate, 
(A1 2 (S0 4 ) 3 ) aluminum sulfate, etc. In a similar manner (HN0 3 ) nitric acid will always form nitrate 
salts containing (N0 3 ) the nitrate radical; (H 2 C0 3 ) carbonic acid will always form carbonate salts 
containing (C0 3 ) the carbonate radical; (H 3 P0 4 ) phosphoric acid will always form phosphate salts 
containing (P0 4 ) the phosphate radical, etc. 

(6) Acids containing oxygen which end in “ous” will form salts the names of which end in 
“ite.” Thus (H 2 S0 3 ) sulfurous acid forms sulfite salts with sulfite radical (S0 3 ); (HN0 2 ) nitrous 
acid forms nitrate salts with the nitrate radical (N0 2 ), etc. 

You will also note that the element naming the acid has a higher valence in compounds ending 
in “ic” and “ate ” and lower valence in compounds ending in “ous” and “ite.” 

(The instructor will outline drill on writing formulas and naming compounds illustrating the 
above.) 


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SUMMARY 


Formula of Acid 

Name of Acid 

Name of Radical 

Formula of Salt 

Name of Salt 

H Cl 

Hydrochloric 

Chloride 



H Br 

Hydrobromic 

Bromide 



H I 

Hydriodic 

Iodide 



H F 

Hydrofluoric 

Fluoride 



h 2 s 

Hydrosulfuric 

Sulfide 



H N0 3 

Nitric 

Nitrate 



H N0 2 

Nitrous 

Nitrite 



h 2 S0 4 

Sulfuric 

Sulfate 



H 2 S0 3 

Sulfurous 

Sulfite 



h 2 S 2 0 8 

Persulfuric 

Persulfate 



h 2 C0 3 

Carbonic 

Carbonate 



H C 2 H 3 0 2 

Acetic 

Acetate 



h 3 P0 4 

Phosphoric 

Phosphate 



h 2 po 3 

Phosphorous 

Phosphite 



H 3 As0 4 

Arsenic 

Arsenate 



H 3 As0 3 

Arsenious 

Arsenite 



H Mn0 4 

Permanganic 

Permanganate 



H 2 Cr0 4 

Chromic 

Chromate 



H 2 Cr 2 0 7 

Dichromic 

Dichromate 



H CIO 

Hypochlorous 

Hypochlorite 



H C10 2 

Chlorous 

Chlorite 



H C10 3 

Chloric 

Chlorate 


* 

H C10 4 

Perchloric 

Perchlorate 



h 3 bo 3 

Boric 

Borate 



H 4 Si0 4 

Silicic 

Silicate 




(Student will fill in for drill.) 


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MIDTERM 


PROBLEM SUMMARY 

(For Drill see Problems in Appendix) 

1. Calculate density when mass and volume given. 

2. Calculate mass when density and volume given. 

3. Calculate volume when mass and density given. 

4. Calculate evolution of calories when data given. 

5. Calculate molecular weight when formula and atomic weights are given. 

6. Calculate percentage composition when formula of compound and atomic weights are given. 

7. Calculate simple formula of compound when percentage of each element is given. 

8. Calculate true formula of compound when percentage of each element and molecular weight 
are given. 

9. Calculate weight of a substance reacting or formed when weight of another substance is 
given. 

10. Calculate weight of liter of various gases from Gram-Molecular-Volume (22.4 liters). 
Instructor will outline drill on these types of problems. 


25 

















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HYDROGEN 


(EIGHTH WEEK) 

25 . (Follow directions carefully. — Two Students.) Fit your wash bottle with thistle tube 
and delivery tube. Place about 10 or 15 grams of zinc in the flask, cover with water, being careful 
to have the lower end of thistle tube under water. Wrap towel around generator. Have instructor 
inspect apparatus. Now slowly pour about 15 cc. of concentrated sulfuric acid down the thistle 
tube. Collect a test tube of the gas by displacement of water and closing with the thumb bring 
it to a flame and light at mouth of tube. Continue testing in this way until the gas burns with 
only a faint explosion. The hydrogen is then free from air and may be safely handled. Collect 
two bottles of the gas by displacement of water, cover with glass squares, and set upside down. 

(a) Remove delivery tube and by means of rubber connection attach a small glass nozzle 
to your generator. Light. Note color and odor of flame (1). To what is the color of 
the flame due (2)? Invert a clean, dry beaker over the burning tip. What happens 
(3)? Write equation (4). 

(b) Holding one bottle of the gas inverted, thrust a burning splinter up into the bottle. 
Does the hydrogen burn (1)? Does it support the combustion of the wood (2)? 

(c) Show the extreme lightness of hydrogen by pouring a bottle of hydrogen up into an 
inverted bottle of air. After a minute test each bottle for hydrogen. (Any undissolved 
zinc may be washed with water and returned to the shelf bottle.) 


26 . (a) In seven test tubes in your rack, place separately small amounts of each of the 
following metals: iron, lead, zinc, magnesium, copper, aluminum, tin. Cover each 
with dilute hydrochloric acid. Do all react (1)? After a minute hold a burning 
splinter in the mouth of each tube. If nothing happens warm the mixture and test 
again. What metals liberate hydrogen with dilute hydrochloric acid (2)? Which 
give none (3)? Record metals in order in which most gas is evolved (4). 

(b) Repeat (a) using only copper, iron, and zinc, adding dilute sulfuric acid. Heat if 
necessary. (1), (2), (3), (4). 

(c) Repeat (b), using acetic acid. (1), (2), (3), (4). 

(d) Repeat (a), using only copper and zinc, adding dilute nitric acid. What acids give 
hydrogen (1)? Write equations for all reactions in (a), (b), (c), and (d), in which 
hydrogen was liberated (2). (See table of activity of metals.) 


27 . Take a clean dry test tubs to stock room and secure a small piece of metallic sodium. 
Add this to test tube half full of water in your test-tube rack. Test the gas (1). Test the water 
in the tube with red litmus (2). What is formed (3)? Equation (4). What other metals will 
act similarly on water (5)? 


28 . To a little powdered zinc or aluminum in a test tube add 5 cc. of concentrated NaOH. 
Caution: Warm and test gas (1). Give equation (2). 


29 . When oxides of metal are heated and exposed to a current of hydrogen gas, the hydrogen 
removes the oxygen and forms water. Write equation for the action of hydrogen on hot copper 
oxide (1). Is this action reversible (2)? What is reduction (3)? What is a hydride (4)? Make a 
list of solid and gaseous reducing agents (5). 


26 




















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WATER 


(NINTH WEEK) 

30 . (Two Students.) Obtain from stock room a condenser, rubber tubing, a 250 cc. side-neck 
distilling flask, a thermometer, and one-hole rubber stopper to fit. Arrange apparatus as directed 
by instructor. Fill flask about half full with tap water and add through a funnel a little colored 
solution such as potassium permanganate. Boil over wire gauze, and collect distillate in clean 
beaker. Note the temperature of the water vapor (1). Why is this not exactly 100 degrees (2)? 
Evaporate a little tap water on a watch glass over a beaker of boiling water. Repeat using dis¬ 
tilled water. Account for the difference in the residues (3). What impurities are not removed by 
distillation (4)? How does distilled water act toward litmus (5)? What other methods than 
distillation may be used to purify water (6)? 


31 . Put a little table salt (sodium chloride) in a test tube containing a little distilled water. 
Warm. Filter into a test tube. Evaporate a little of the clear filtrate in your evaporating dish. 
Is there a residue (1)? If so, what is it (2)? Has a chemical change taken place (3)? Define a 
solution (4). (See table of solubility in Appendix.) 


32 . Place about two grams of each of the following substances in test tubes and boil with 
10 cc. of water: sodium chloride, potassium dichromate, calcium sulfate, calcium carbonate. 
In each case set aside, cool, and filter. Are these substances more soluble in hot water (1)? Evapo¬ 
rate a little of each of the filtrates on watch glasses over beaker of boiling water and record the 
comparative solubilities as shown by residues (2). How determine whether a substance is soluble 
in water (3)? (See table of solubility in Appendix.) 


33 . Warm a few small crystals of each of the following substances separately in dry test tubes: 
barium chloride, potassium nitrate, magnesium sulfate, potassium dichromate, and copper 
sulfate. Is moisture driven off (1)? Is this water of crystallization (2)? Write formulas of these 
compounds, showing water of crystallization where present (3). 


34 . Place a little blue vitriol (cupric sulfate) in an evaporating dish and heat gently over a 
flame until a white product remains. Has a chemical change occurred (1)? Place a small portion 
of this white product in a test tube and add water and boil. What has happened (2)? Pass steam 
over another portion. Do you notice any change (3)? Explain and write equation (4). 

t 


35 . Heat a gram of common salt (sodium chloride) in a test tube. Does salt contain water of 
crystallization (1)? What is decrepitation (2)? 


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36 . Heat about a gram of common sugar in a test tube. Is water driven off (1)? Is this water 
of crystallization (2)? What is the residue (3)? When water is added to this residue is the 
original sugar produced (4)? 


37 . To about a gram of sugar in a test tube add sufficient concentrated sulfuric acid to cover. 
Note change in the sugar (1). Compare the residue with that obtained in (36). What becomes of 
the water removed from the sugar (2)? What is water of constitution (3)? W hat is a dehydrating 
agent (4)? 


38. As a summary, water may be found with chemical substances either physically due to 
adhesion to the particles or chemically combined as water of crystallization or water of constitu¬ 
tion. Compounds containing water of crystallization are called hydrates. Such water may be 
removed and again replaced. When water of constitution is removed from compounds it may 
be replaced as in the case of calcium hydroxide or sulfuric acid or it may be impossible to replace 
the water as with sugar and starch. Give other illustrations of hydrates and compounds con¬ 
taining water of constitution (1). 


28 







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LAWS OF CHEMICAL COMBINATION 


(TENTH WEEK) 

39. Combining Weight or Hydrogen Equivalent of Metals. — Fit a wash bottle with a one-hole 
stopper through which is inserted a short piece of glass tubing bent to a right angle. This is at¬ 
tached by short rubber connection with pinchcock to a second and then to a third wash bottle. 
These wash bottles are fitted with two-hole stoppers through which an inlet tube bent at right 
angle extends through the stopper and the outlet tube bent at right angle extends to the bottom 
of the bottle and is in turn attached to the next wash bottle by a rubber connection. The final 
rubber outlet tube closed by a pinchcock is placed in a dry, one-liter beaker. The last two wash 
bottles and connecting tubing from one pinchcock to the other are completely filled with water, 
driving out all air bubbles. Your instructor will give you the exact weight of a definite length of 
magnesium ribbon. Measure with a meter stick the required length to give 0.3 gm. of the ribbon. 
Into the first wash bottle by means of a funnel introduce 15 cc. of concentrated hydrochloric acid 
and 15 cc. of water. Lay the wash bottle on its side, cut the magnesium ribbon in short lengths, 
and place in the neck of wash bottle. Do not allow the ribbon to come in contact with the acid. 
Place stopper and delivery tube in wash bottle. Open the pinch clamps and slowly tilt the appa¬ 
ratus upright so that pieces of the ribbon may drop in the acid. When all of the magnesium has 
been decomposed adjust the levels of the water in the beaker and bottle, and clamp tubes. Observe 
the temperature of the laboratory and the barometric reading, making correction for tension of 
aqueous vapor for latter. Reduce the volume of hydrogen represented by the volume of water in 
the beaker (measure in graduated cylinder) to standard conditions, and from the weight of 22.4 
liters (2.016 gms.) calculate the weight of this volume of hydrogen. From this weight of hydrogen 
obtained, calculate the equivalent or combining weight of magnesium that is exactly equivalent to 
1.008 gms. of hydrogen. 

(a) Repeat the above experiment with 0.2 gm. aluminum wire. 

(b) Repeat the above experiment with 0.2 gm. iron wire. 

(c) Repeat the above experiment with an unknown metal or alloy furnished by instructor. 


DATA 



Mg 

Al 

Fe 

Unknown 

Weight of metal used. 

Volume of hydrogen (H 2 0 displaced).. . 

Temperature of water. 

Barometric pressure. 

Aqueous pressure. 

Partial pressure of hydrogen. 

Volume of hydrogen at S. C. 

Weight of hydrogen. 

Equivalent or combining weight. 




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THE THREE STATES OF MATTER AND THE GAS LAWS 

(ELEVENTH, TWELFTH, AND THIRTEENTH WEEKS) 

(Extra laboratory time will be devoted to theory) 

41. Boyle’s Law 

(a) By means of a rough sketch show how barometric readings are taken. If the reading 
is 740 mm. represent mercury levels on your sketch (1). Readings may vary on dif¬ 
ferent days. 

(b) What is standard pressure (1)? What is one atmosphere or 15 pounds per square inch 
expressed in millimeters of mercury (2) ? 

(c) If the density of mercury is 13.6, what would be the length of water column corre¬ 
sponding to 760 mm. at mercury (1)? 

(d) 100 liters of gas were collected at each of the following pressures: Calculate the 
volume of gas in each case at standard pressure: 720 mm., 840 mm., 755 mm., 780 
mm. (1), (2), (3), (4). 

The following procedure wiU eliminate necessity of memorizing a formula: 

1. Read your problem carefully. 

2. Remember that more mm. of mercury represents more pressure and consequently 
a contraction of volume. 

3. Ask two questions: first, is the pressure getting greater or is it less? Second, 
what will therefore happen to the volume? 

4. If you wish the volume to increase, multiply the volume given by the fraction in 
which the greater pressure is the numerator; if the volume is to decrease, the 
smaller pressure will be the numerator of the fraction. 


42. Charles’ Law 

(a) By means of a rough sketch represent three thermometer scales — Centigrade, 
Fahrenheit, and Absolute, showing relation between freezing point and boiling point 
of water on each (1). 

(b) What is standard temperature (1)? Why are all temperatures converted to absolute 
before correcting volumes of gases (2)? Convert the following to absolute degrees: 
32° F., 100° C., 10° C., 215° F. (3), (4), (5), (6). 

Note. —For conversion of Centigrade to Fahrenheit or reverse, the following system 
may be used: 

Fahrenheit to Centigrade 5/9 (F.° — 32) = C.° 

Centigrade to Fahrenheit 9/5 C,° + 32 = F.° 

If you do not care to remember this conversion formula proceed as follows: 

1. Represent the two thermometer scales Centigrade and Fahrenheit by parallel 
vertical lines. Draw two horizontal lines across these — one representing freezing 
point, and the other, boiling point on both scales. The distance between these 
two points is 100° on the Centigrade and 180° on the Fahrenheit scale. Thus a 
Fahrenheit degree is 100/180 or 5/9 as large as a Centigrade degree. 

2. Assume a reading of 20° C. to convert to F. Represent on your diagram. 20° on 
the Centigrade scale will give more than 20° or 20 X 9/5 = 36°; thus the reading 
is 68 ° F. 

3. Assume a reading of 77° F. to convert to C. Represent on your diagram. Of this 
77°, we have 32° from 0° to freezing point, 32° F.; therefore we have only 45° F. 


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above the freezing point. Since a Fahrenheit degree is only 5/9 as large as the 
Centigrade, our reading will now be 45° X 5/9= 25° C. 

(c) 100 liters of gas were collected at each of the following temperatures; calculate the 

volume of gas in each case at standard temperature: 20° C., — 5° C., 100 0 C., 

40° C., - 15° C. (1), (2), (3), (4), (5). 

The following procedure will eliminate necessity of memorizing a formula: 

1. Read your problem carefully. 

2. Convert the Centigrade or Fahrenheit degrees to Absolute, substituting the new 
figures. 

3. Remember that the higher the temperature the more the gas will expand. 

4. Ask two questions: first, is the temperature increasing or is it decreasing? Second, 
how will this affect the volume of gas? 

5. If you wish the volume to increase, multiply the volume given by the fraction in 
which the greater absolute temperature is the numerator; if the volume is to 
decrease, the smaller Absolute temperature will be the numerator of the fraction. 


43. Charles’ Law and Boyle’s Law Combined. — Correction is made for both temperature and 
pressure change as before, combining the two problems as one. If the gas is collected over 
water the aqueous tension in millimeters of mercury must be subtracted from the observed 
pressure before representing the pressure fraction correction of volume. 

(a) 100 liters of gas were collected over water at each of the following temperatures and 

pressures; calculate the volume of gas in each case at the new temperatures and 

pressures. 

(1) at 20° C. and 750 mm. to 10° C. and 800 mm. 

(2) at 15° C. and 770 mm. to 25° C. and 750 mm. 

(3) at 25° C. and 740 mm. to Standard Conditions. 

(4) at 10° C. and 780 mm. to 20° C. and 800 mm. 

(5) at Standard Conditions to 30° C. and 740 mm. 


32 


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SOLUTIONS 


(FOURTEENTH AND FIFTEENTH WEEKS) 

Note. — Extra laboratory time will be devoted to theory. 

44 . Fill a test tube one-fourth full of sodium acetate crystals, add about 1 cc. of water, and 
apply heat carefully till all of the salt has dissolved. Loosely stopper the tube and allow to cool. 
Now add a small crystal of sodium acetate. Explain the result (1). Distinguish between an un¬ 
saturated solution and a supersaturated solution (2). 


45 . (a) Place 50 cc. of water in a beaker. Record the temperature of the water (1). Now add 
two grams of CaCl 2 and record temperature (2). Heat to boiling, record temperature 
of the boiling solution (3). Compare this temperature with the temperature of boiling 
distilled water (4). 

(b) Repeat, using NH 4 C1. (1), (2), (3), (4). 

(c) What is exothermal heat of solution (1)? What is endothermal heat of solution (2)? 
Is the boding point of the solution higher or lower than the boding point of pure water 
(3)? Explain (4). 


46 . (a) To 5 cc. of water add a few drops of indigo solution, 2 grams of powdered charcoal, and 
heat to boding. Fdter. Does the charcoal remove any of the coloring material (1)? 
(b) To 10 cc. of water add one drop of ammonium hydroxide, 2 grams of powdered 
charcoal and water. Filter. Compare odor before and after addition (1). Is char¬ 
coal used as a fdter for drinking water? Why (2)? Why are charcoal tablets given 
for indigestion (3)? Why used in gas masks (4)? 


47 . Test the solubility of carbon bisulfide in water, chloroform in water, ether in water, alcohol 
in water, and kerosene in water, using only a few drops of water in each case and about 5 cc. of the 
other liquid. Shake well in each case and note whether solution has occurred or whether there 
are two layers of liquid. The presence of smad traces of water may be detected by adding a small 
crystal of potassium permanganate. If water is present a pink solution wid result. Make a fist 
of liquids soluble in each other and liquids insoluble in each other (1). 


33 
































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48 . (a) Place a single crystal of iodine in a test tube, add 5 cc. of carbon bisulfide, shake, 
and note color of solution (1). 

(b) Place a single crystal of iodine in a test tube, add 5 cc. of water, shake, and note 
whether any iodine has dissolved (1). Does heat increase solubility (2)? 

(c) Into a clean test tube pour a little of this aqueous solution of iodine, add a little car¬ 
bon bisulfide, shake, and note results (1). Compare solubility of iodine in water and 
in carbon bisulfide (2). 

(d) Repeat (a), substituting a crystal of potassium iodide for carbon bisulfiide (1). 
How would you make an aqueous solution of iodine (2)? What is tincture of iodine 
(3)? Define solvent, solute, saturated solution (4). Make a list of as many factors 
as you can think of which affect solubility of substances (5). 


49 . Use the following: zinc oxide, sodium oxide, calcium oxide, iron oxide, copper oxide, and 
sand (silicon dioxide). Add a little of each to a few cc. of water in a test tube and boil if it does not 
dissolve in the cold. Test each solution with litmus paper (1). What other oxides (third week) 
have you treated with water (2)? What did their solutions do to litmus (3)? Classify the oxides 
you have studied as follows: (a) oxides of metals; (b) oxides of nonmetals; (c) oxides which 
dissolve in water; (d) oxides which are insoluble in water; (e) oxides which form acids with water; 
(f) oxides which form bases with water (4). 


34 





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IONIZATION 

(SIXTEENTH WEEK) 


50. Electrical Conductivity. — One wire of an extension cord to a lighting circuit is cut and a 
socket and lamp globe attached to these cut ends. The two wires at the end of the extension are 
stripped of insulating material for a short distance. These two bare ends are insulated from each 
other by running through glass tubing or a two-hole stopper, which is bound in place by electric 
tape or sealing wax. When these bare ends are put in contact with a conductor the lamp globe 
will light up with a brightness somewhat proportional to the conductivity. After each test rinse 
the bare ends of wire with distilled water. Using clean evaporating dishes or beakers as containers, 
test several of each of the following substances: distilled water, dry salts, dry organic substances, 
concentrated acids, dry sodium hydroxide or potassium hydroxide, solutions of acids, bases, salts, 
and organic substances. 

Record all substances under the heading “conductors” and “non-conductors” (1). Explain 
all results (2). 


51. Equilibrium in Solution. 

(a) Dissolve about 1 gm. of cupric bromide in the least possible volume of water. What is 
the color of the solution (1)? Add water, a small quantity at a time, until solution 
becomes a clear green. Pour off half of the solution and dilute it further. What is 
the final color (2)? How can you account for the intermediate green (3)? To the 
remaining half of the green solution add crystals of potassium bromide or sodium 
bromide. What change in color is noticed (4)? How do you explain this (5)? 

(b) Set up six clean test tubes in your rack and fill about half full as follows: 

(1) Ammonium thiocyanate (NH 4 CNS) solution. 

(2) Ferric chloride (FeCl 3 ) solution. 

(3) Mixture of 1 cc. of (1) and (2) diluted to 50 cc. 

(4) This mixture (3) and more NH 4 CNS. 

(5) This mixture (3) and more FeCl 3 . 

(6) This mixture (3) and solid NH 4 C1. 

Compare colors and record any changes in color which occur when reagents are added (1). 
Give equilibrium equation involved (2). Explain from an ionic viewpoint the action in each tube 
(3). What ions are common in the last three tubes (4)? 


35 






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62. Conditions Affecting Equilibrium in Solution. 

(a) Use the same apparatus as in Expt. 50. To a clean beaker half filled with distilled 
water add drop by drop enough dilute hydrochloric acid to cause a very faint glow 
of the lamp globe when the bare ends of wire are immersed. In another clean beaker 
half filled with distilled water add drop by drop enough dilute sodium hydroxide 
solution to produce a similar faint glow of the globe when the wire ends are immersed. 
In a third clean beaker pour equal quantities of the solution from each of the two 
beakers and test with the bare ends of wire. Is there a glow of the lamp globe (1)? 
Write the ionic equation involved (2). What substance is formed which you found in 
Expt. 60 to be a non-conductor or non-ionizing (3)? 

(b) Repeat (a), using instead of the hydrochloric acid and sodium hydroxide, one beaker 
with sufficient barium hydroxide and one beaker with sufficient sulfuric acid to 
produce faint glow. Answer the same questions (1), (2), (3). 

(c) In a chemical reaction in which a gas is liberated would you expect an equilibrium to 
be reached (1)? Summarize the conditions under which unstable equilibrium will 
exist (2). 


63. The Relationship of Dilution to Ionization. — This experiment requires a wooden V- 
shaped trough approximately 10 inches deep, 21 inches across the top, and about 8 inches long. 
This must be water-tight, and is further waterproofed by covering the inside with melted paraffin. 
Two triangular-shaped copper sheets are cut to fit and hung over each end of the trough. A screw- 
tap connection is soldered to each strip of copper. One wire leads from a terminal of a battery to 
one connection on the trough. The other battery wire leads through an adjustable rheostat and 
an ammeter to the other connection on the trough. A voltmeter is shunted across the circuit 
from the two battery wires after the one has passed through the rheostat. The ammeter should 
have a capacity of about two amperes, the voltmeter a capacity of ten or fifteen volts, and the 
rheostat about thirty ohms. 50 cc. of a saturated solution of copper nitrate and 50 cc. of dis¬ 
tilled water are poured in the trough in contact with the copper sheets. Adjust the rheostat 
until a current of 0.1 to 0.2 amperes is passing. Record this reading in amperes and the reading in 
volts noted on the voltmeter. Now add 100 cc. of distilled water to the solution in the trough. 
Adjust the rheostat until the voltage is the same as the first reading and note the new reading 
on the ammeter. Continue in this manner, doubling the volume of solution in the trough with 
each dilution, keeping the voltage constant and recording the ammeter readings. Tabulate re¬ 
sults as follows: 


Volume 

Voltage ( Constant) 

Current (Amps) 

Ratio of Current to 
Maximum Current 

Current per 100 cc. 

100. 





200. 

400. 

800. 

1600. 

3200. 

6400. 

Etc. 



Interpret your results (11. 


36 

























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54. Migration of Ions. — Saturate a 5% solution of agar-agar with crude sodium chloride or 
potassium dichromate. When the material has solidified in a U-tube cover with a solution of 
sodium chloride and by means of two platinum electrodes immersed in the sodium chloride solu¬ 
tion allow a direct current to pass through the mixture. If the experiment is allowed to run for 
about thirty minutes a movement of the colored ions can be observed. If available, cupric chro¬ 
mate can be used with even better results. The agar-agar solution is used simply to prevent 
movement of the colored salt during the experiment. 

Draw diagram of apparatus and explain results. 


* 


37 










NEUTRALIZATION, TITRATION, AND VOLUMETRIC CALCULATION 

Since neutralization is the combination of hydrogen and hydroxide ions to form the non¬ 
ionizing substance, water, we may determine the amount of acid or base in an unknown by neu¬ 
tralizing it with a solution containing a known amount of acid or base. Such a solution is called a 
standard solution. The liter or 1000 cc. is the common standard of volume. If a solution is of 
such strength that it contains the molecular weight in grams of the substance in a liter of the solu¬ 
tion, we call this a molar solution. If the solution contains one hydrogen equivalent in a liter of 
the solution it is called a normal solution. Thus normal HC1 requires 36.468 gms. of HC1 per liter 
to contain 1.008 gm. of hydrogen. In the same way normal NaOH requires 40.008 gms. of NaOH 
per liter to contain 23 gms. of sodium, which is the equivalent of 1.008 gms. hydrogen. Since 
hydrogen carries a positive charge in solution, we can obtain the equivalent weights required for 
normal solutions by dividing the molecular weights of the compounds concerned by the total 
number of positive charges. Thus normal H 2 SO 4 will contain 49.043 gms., normal H3PO4 will 
contain 32.685 gms., normal A 1 2 (S 04)3 will contain 57.02 gms., per liter of the solution. A tenth 
normal (N/10) solution of these will contain one-tenth of the weights per liter required for the 
normal solution. 


55. Calculate weights per liter required for each of the following: 

(1) Tenth normal HC1. 

(2) Fifth normal NaOH. 

(3) -Five normal HC1. 

(4) Half normal H2SO4. 

( 5 ) Tenth normal H3PO4. 


56. Calculate the grams per cubic centimeter (called the titer) of each of these in (55). (1), 
(2), (3), (4), (5). 


57. Secure from the storeroom two burettes. Clean, rinse with distilled water and rinse with 
the solution to be measured, and attach by clamps to your iron stand. Fill one burette with some 
desk sulfuric acid and the other with some sodium hydroxide of known strength (furnished by the 
instructor). Run 20 cc. of the acid from the burette into a clean beaker, add a few drops of phe- 
nolphthalein solution and about 20 to 30 cc. of distilled water, and then slowly add the sodium 
hydroxide from the other burette, stirring between drops, until a faint pink color is imparted to 
the solution. From the figures obtained calculate (1) the titer of the acid and (2) the normality 
of the acid. 
















' 





' 



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. 

' 

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58. Repeat (57), using some commercial vinegar and calculate (1) the titer of acid, (2) the 
normality of the acid, and (3) the percentage of acid in the vinegar. 

Sample Calculation. — Assume in (67) that 30 cc. of N/10 NaOH neutralized the 20 cc. of 
unknown strength H2SO4. To solve for normality and titer of the acid, either of the following 
methods may be used: 

Method (1). — Since normal NaOH contains 40.008 gms. per liter, N/10 will contain 4.0008 
gms. per liter or .004 + gms. per cc. Thus 30 cc. of this NaOH will contain 30 X .004 gms. of 
NaOH. The determination of titer of H 2 S0 4 now becomes a simple weight problem: 

H2SO4 -(- 2 NaOH = Na 2 S0 4 d - 2 H 2 0 
98 80 

(20cc. XX) (30cc. X .004) 

80 :98 :: 0.120 :20X 

X = .00735 gms. H 2 SO 4 per cc. (Titer). 

Normal H 2 S 0 4 contains 49.043 gms. per liter or 0.049043 gms. per cc. This H2SO4 is therefore 
.00735/.049043N or N/6.67 (Normality). 

Method (2). — Equal volumes of same strength acids and bases always neutralize. In case 
equal volumes are not used but the strength of one is known, always compare the unknown with 
the known. Thus in this problem, from the volumes given we see that the unknown H 2 SO 4 is 
30/20 stronger than the NaOH. Since the NaOH is N/10, the acid must be 30/20 X N/10 = 
30N/200 or N/6.67 (Normality). If normal H 2 S0 4 contains 49.043 gms. per liter or .049043 gms. 
per cc., this acid will contain .049043/6.67 or .00735 gms. H 2 SO 4 per cc. (Titer). 


39 








SULFUR AND ITS COMPOUNDS 

(SEVENTEENTH AND EIGHTEENTH WEEKS) 

59. (a) To a gram of powdered sulfur in a test tube add a few cc. of carbon bisulfide and 
shake. Filter off some of the clear solution into a watch glass and allow evaporation 
to take place in air. Examine crystals with a lens. Are they transparent (1)? Draw 
a rough sketch showing their shape (2). 

(b) (Two Students.) Half fill a porcelain crucible with sulfur set on a ring stand and 
heat gently, until all the sulfur has just melted. Remove flame and allow the molten 
sulfur to cool until a crust forms on the surface. Puncture this crust and pour out 
the molten sulfur in a beaker of cold water, saving beaker and contents for (c). 
With a lens examine the crystals adhering to the side of the crucible and compare with 
those obtained in (a). Are they transparent (1)? Make sketch (2). 

(c) Remove the solidified sulfur from the beaker of cold water, dry between filter paper, 
and test solubility in carbon bisulfide (1). 

(d) Heat 5 gms. of sulfur in a test tube until it becomes brown and viscous and then 
boils. Pour into a beaker of cold water, remove solidified sulfur from the beaker, 
dry between filter papers, and test solubility in carbon bisulfide (1). 


60. Prepare sulfur dioxide by gently heating some concentrated sulfuric acid in a test tube 
with a few copper shavings. By what other method has this gas been previously prepared (1)? 
Test the gas with moistened litmus, moist paper containing writing ink, moist newspaper, and 
moist calico. Are the colors affected (2)? 


61. (a) Test an aqueous solution of sulfur dioxide (shelf) with blue litmus. Is the gas com¬ 
bined with the water or merely dissolved in it (1)? Give reason (2). Boil a little of 
the solution. Does odor persist (3)? How explained (4)? 

(b) To 5 cc. of the solution add a few drops of barium chloride. Is there a precipitate (1)? 
Is it soluble in hydrochloric acid (2)? Repeat this test, first adding a little concen¬ 
trated nitric acid and boiling before adding barium chloride. Do you note any differ¬ 
ences (3)? How explained (4)? 

(c) Pass a little hydrogen sulfide gas into a solution of sulfur dioxide in water. What is 
the precipitate (1)? Equation (2). 


40 




























62. (a) On the ring of your iron stand over wire gauze set up a wash bottle fitted with two-hole 
stopper through which a thistle tube extending to the bottom and an outlet tube bent 
at right angle are inserted. This flask is to be used for generation of sulfur dioxide as 
in Expt. 60. Fit a small Erlenmeyer flask with one-hole stopper and outlet tube to be 
used for generation of nitric oxide using copper shavings and concentrated nitric acid. 
The outlet tubes from these flasks extend just through a four-hole stopper which is 
fitted to a wide-mouthed clear glass bottle. Through the third hole of this stopper 
a glass tube bent at right angle extends down into a small amount of water in the 
bottom of the bottle. An outlet tube bent at right angle extends through the fourth 
hole of the stopper and leads to a beaker of water. An aspirator bulb is attached to 
the tube leading to the bottom of the bottle, by means of which air can be bubbled 
through the water. Warm the contents of the wash bottle and conduct the sulfur 
dioxide to the bottle, where it mixes with the red nitric fumes and air from the 
aspirator bulb. Excess of sulfur dioxide destroys the red color which will reappear 
when more air is admitted. What are the white crystals which form on the walls of 
the bottle (1)? Are they soluble in the water (2)? Test the contents of the bottle 
after disconnecting the apparatus, in the same manner as you test sulfuric acid in 
Expt. 62 (b). Write all equations involved in the preparation and test of this acid (3). 

(b) Pour a little concentrated sulfuric acid slowly into a half test tube of water. Is 
there a temperature change (1)? Test this solution with a few drops of barium 
chloride. Is this precipitate soluble in hydrochloric acid (2)? The formation with 
barium chloride of a precipitate which is insoluble in hydrochloric acid is a good test 
for sulfuric acid or the sulfate ion. Try this test on some tap water (3). 

(c) What is the result of dipping a wooden splinter in concentrated sulfuric acid (1)? 
Of pouring the concentrated acid on dry sugar or starch (2)? Explain (3). Why is 
sulfuric acid used in preparing other acids (4)? What other acid may be used (5)? 


63. Make a hydrogen sulfide generator using a 250 cc. Erlenmeyer flask, one-hole rubber 
stopper, and delivery tube. Have at hand a dry bottle which is to be filled with gas. In test tubes 
in your rack have small quantities of each of the following solutions: copper sulfate, lead nitrate, 
cadmium sulfate, zinc sulfate, magnesium sulfate, and sodium chloride. Now to a few pieces 
of iron sulfide in your flask add a little dilute sulfuric or hydrochloric acid, adding more acid from 
time to time as needed. (Caution: Do not inhale the gas as it is poisonous.) 

(a) Expose a piece of moist litmus paper to the gas. What is the reaction (1)? Dip a 
piece of filter paper in some lead nitrate solution and expose to the gas. Give formula 
of the black deposit (2). 

(b) Fill the dry bottle with the gas by downward displacement of air until the gas is 
ignited by a flame at the mouth of the bottle. What is the deposit on the sides of the 
bottle (1)? Connect a glass nozzle to the delivery tube and ignite the gas at the tip. 
What is the color and odor of the flame (2)? Hold a cold porcelain dish in the flame. 
What is the deposit (3)? Give equation for this reaction (4). 

(c) Bubble some of the gas through each of the solutions in your rack, noting the color 
of the precipitate obtained. Give equation in each case (1). 

(d) Pass some of the gas into some water in a beaker. Compare the odor of this solution 
with the odor of natural sulfur waters (1). Pour a little of the solution in a test tube 
and boil. Does the odor disappear (2)? As long as any of the gas is being given off, 
filter paper moistened in lead nitrate solution and held in the mouth of the tube will 
be darkened. Drop a silver coin in some of the solution of the gas in water. Explain 
the results (3). 

Note. — Wash any remaining pieces of iron sulfide in your Erlenmeyer and return to 
the shelf bottle. 


41 







SECOND TERM 


THE ATMOSPHERE, NITROGEN AND ITS COMPOUNDS 

(FIRST AND SECOND WEEKS) 

64. Float a flat cork on water in a pneumatic trough. Place a small amount of red phos¬ 
phorus on the cork, ignite by hot wire or flame, and carefully invert a Mason jar over the cork, into 
the water. What are the white fumes (1)? When these fumes have been absorbed by the water, 
paste a sticker at the point to which the water has risen in the jar, being sure that jar is lowered 
until inside and outside water levels are the same. What has been removed from the air in the jar 
(2)? The contents of the jar may be assumed to be about 98 per cent nitrogen. Measure the 
volume of the jar using your graduated cylinder, and then from the volume occupied by the ni¬ 
trogen, calculate approximately the percentage of nitrogen in atmospheric air (3). What is the 
approximate percentage of oxygen (4)? 


66. Oxygen may be removed from a mixture of gases by absorption in an alkaline solution of 
pyrogallic acid. Pour about 3 cc. of alkaline pyrogallic acid in a test tube, close mouth of tube 
with thumb, and shake thoroughly. Invert tube in a tank of water, remove thumb, lower tube 
till water levels on inside and outside of tube are the same, and measure and calculate percentage 
of oxygen and nitrogen in air as in (64). 


66. To prove the presence of carbon dioxide in air allow a little limewater to stand in an open 
beaker. What is the scum formed (1)? Blow your breath, which contains carbon dioxide, over 
some limewater in a beaker. Is a similar result obtained (2)? Give equation (3). 


67. In a test tube place a mixture of 3 grams of sodium nitrite and 2 grams of ammonium 
chloride. Warm gently and test gas with a burning splinter. Does the gas support combustion (1)? 
Is it combustible (2)? What is its color and odor (3)? Write equation for this reaction (4). 


68. In a test tube place 2 grams of powdered lime and ammonium chloride, and warm 
gently. Note the odor of the gas (1). Test the gas with moistened red and blue litmus paper. 
The presence of what ion is indicated (2)? Hold the stopper of your hydrochloric acid in the gas. 
What are the fumes (3)? Is the gas soluble in water (4)? Give equation for this reaction (5). 


69. Neutralize a little ammonium hydroxide with dilute sulfuric acid using phenolphthalein 
as an indicator. Evaporate a little of this solution to dryness in your evaporating dish. What is 
the residue (1)? Write ionic equation (2). Invert a funnel over the dry residue on the dish, closing 
the stem of the funnel with a paper plug. Heat dish gently and notice the sublimate. Is this the 
same as the original (3)? 


42 










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. 



















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70. In a test tube containing 2 cc. of ammonium chloride solution pour an equal amount of 
sodium hydroxide solution and warm gently. Note odor (1). Test with moistened litmus (2). 
Give equation (3). This evolution of ammonia by the action of a strong base on an ammonium 
salt is a definite test for the presence of the ammonium radical. 


71. In a test tube place 2 grams of sodium nitrate crystals, add 2 cc. of concentrated sulfuric 
acid, and when the acid has saturated the mass of nitrate, warm gently. Write equation for the 
reaction (1). Is this reaction reversible (2)? Do not breathe these fumes or allow them to come 
in contact with the flesh. Blow moist air from the breath across the mouth of the tube. What 
are the fumes (3)? Hold an ammonia stopper at the mouth of the tube. What are the fumes (4)? 
Test the gas with moistened litmus paper (5). 


72. Moisten a glass rod with concentrated nitric acid and touch to your finger nail or calloused 
flesh of your hand. After about thirty seconds wash off with water. Note the color of the flesh (1). 
This formation of yellow Xanthoproteic acid-is a test for the presence of protein. Try the same 
test on samples of cotton and woolen cloth. Is the color similar and if so, has the same reaction 
taken place (2)? 


73. Drop a piece of copper turnings into 2 cc. of fairly concentrated nitric acid and note the 
brown fumes of nitrogen tetroxide. Is this a reaction of the molecular nitric acid or of the nitrate 
radical upon copper? Determine by adding copper to each of the two test tubes — one contain¬ 
ing a little sodium nitrate in solution, and the other a little sodium nitrate and concentrated 
sulfuric acid (1). 


74. Warm the following metals in a little dilute nitric acid, noting color of the gas liberated: 
magnesium, zinc, tin, copper (1). Is hydrogen liberated (2)? Is the nitric acid decomposed (3)? 
Does oxidation occur (4)? Give equations (5). 


75. Repeat (74), substituting concentrated nitric acid. Answer the same questions. 


76. To 1 cc. of sodium nitrate solution in a test tube add 3 cc. of water, 3 cc. of ferrous 
sulfate solution, shake, incline tube, and carefully pour a little concentrated sulfuric acid down 
the inside of the tube. The brown ring at the junction of the heavy sulfuric layer and the mixture 
is a delicate test for the presence of the nitrate radical. 


43 







THE CHLORINE FAMILY 


(THIRD AND FOURTH WEEKS) 

77. Chlorine.— (Caution: This gas is poisonous.) 

(a) In test tubes try the action of a little concentrated hydrochloric acid on each of the 
following, heating tube gently if necessary: manganese dioxide, lead dioxide, po¬ 
tassium permanganate, potassium dichromate. Will all compounds containing oxygen 
oxidize hydrochloric acid (1)? Try litharge (lead oxide) and magnesium sulfate (2). 

(b) Test the gas evolved from the test tubes, as follows, recording results and giving 
equations: 

1. Moist starch potassium iodide paper. 

2. Drop a pinch of powdered antimony into one tube. 

3. Moist colored calico. 

4. Moist paper containing writing ink, also paper with news print. 

5. Filter paper soaked in turpentine (do not hold in hand). 

(c) Write balanced equations for reactions in (a) (1). Why is moisture necessary to cause 
chlorine to bleach (2)? What is chlorine water chemically (3)? 


78. Bromine. — (Caution: These fumes are poisonous.) 

(a) Repeat Expt. (77) (a), substituting hydrobromic acid in place of hydrochloric. Note 
the color and odor of the gas (1). Does it have the same bleaching properties as 
chlorine (2)? 

(b) To a little bromine water (shelf) in a test tube add a little carbon bisulfide and shake 
(1). Repeat, adding a little chloroform to the bromine water (2). Is bromine more 
soluble in these than in water (3)? 

(c) Shake a little potassium bromide solution in a test tube with carbon bisulfide. Is 
bromine shown (1)? Now add a few drops of chlorine water in the tube and shake. 
Explain (2). Write equation for reaction (3). 


79. Iodine. 

(a) In a test tube place a mixture of 1 gram of potassium iodide or sodium iodide, and 2 
grams of manganese dioxide. Add about 3 cc. of dilute sulfuric acid and warm gently. 
Note color of the fumes (1). What is the deposit on the cool portions of the tube (2)? 

(b) Make an aqueous solution of iodine by warming a crystal of iodine in a half test tube 
of water. Half fill three test tubes with starch solution, and treat as follows: 

1. To the first add a few drops of the iodine solution. Note color (1). 

2. To the second add a few drops of chlorine water. Is there any action (2)? 

3. To the t hir d add a few drops of potassium iodide solution. Is there any action (3)? 
To this tube now add a few drops of chlorine water. What has happened (4)? 
Give equation (5). 

(c) To a half test tube of aqueous solution of iodine add 2 cc. of carbon bisulfide and 
shake. What is the colored layer (1)? 

(d) A concentrated aqueous solution of iodine may be made by adding a crystal of po¬ 
tassium iodide to a crystal of iodine, adding a small amount of water and shaking. 
Why does iodine sometimes blister the skin (1)? Why.is it a good antiseptic for cuts 
and wounds (2)? 


44 















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80. Hydrofluoric Acid. — (Caution: The gas and its solutions are very corrosive and must not 
be inhaled or brought in contact with the skin.) 

Warm very gently a small piece of paraffin on a glass square, covering the entire surface with a 
thin coating of wax. When cold cut some letters or diagram through the wax. Place in a lead 
dish (stock room) a little powdered calcium fluoride and add enough concentrated sulfuric acid 
to make a paste. Cover dish tightly with the waxed side of the glass square, warm the bottom of 
the dish in a low luminous flame for an instant, then set aside for an hour. Remove the paraffin 
and examine the glass. What has happened (1)? Write equations for reactions involved (2). 
What kind of bottles are used for storing hydrofluoric acid (3)? 


81. Hydrogen Chloride and Hydrochloric Acid. — Fit an Erlenmeyer flask with a two-hole 
rubber stopper, thistle tube, and delivery tube. To 10 or 15 grams of common salt in the flask 
add 20 cc. of concentrated sulfuric acid through the thistle tube. Mix by motion of flask and 
warm gently. Collect two dry bottles of the gas by downward displacement of air, and cover 
tightly with glass plates. Allow the gas to flow down into a beaker of water without allowing the 
delivery tube to come in contact with the water. 

What is the color and odor of hydrogen chloride gas (1)? What causes the fumes when the gas 
is exposed to the air (2)? What is the effect of blowing your breath through the gas (3)? Of hold¬ 
ing stopper of ammonium hydroxide bottle in the gas (4)? What are the fumes in each case (5)? 
Equations (6). How does the gas affect moist litmus paper (7)? Does it have any action on the 
dry litmus (8)? Test the water with litmus. What has been formed (9)? 


82. Hydrobromic Acid. 

(a) In a test tube containing a few crystals of potassium bromide add a little sulfuric 
acid and warm gently. What are the white fumes (1)? What is the reddish brown 
vapor (2)? If this method in Expt. (81) gives the hydrohalogen acid, why is it not 
advisable in the preparation of hydrobromic acid (3)? Give equation for the labora¬ 
tory preparation of pure hydrobromic acid (4). 

(b) Recall action of an oxidizing agent on hydrobromic acid. (Expt. 78) (a) (1). 

(c) Repeat (78) (c), substituting hydrobromic acid for potassium bromide. Give equa¬ 
tion (1). 


83. Hydriodic Acid. — Repeat (82) (a), substituting potassium iodide for the potassium bro¬ 
mide. Ans wer the same questions and give equation. 


45 



' 










, 

■ 

■ 












. 

' 




84. Summary. 

(a) Detection of free halogens. 

1. Try the action of silver nitrate solution on solutions of chlorine, bromine, and 
iodine. Note color of each precipitate (1). Test the solubility of each precipitate 
with excess of ammonium hydroxide (2). 

2. Add carbon bisulfide to solutions of chlorine, bromine, and iodine. Note char¬ 
acteristic color of each after shaking (3). 

(b) Detection of combined halogens. 

1. Repeat (a), using sodium chloride, potassium bromide, and potassium iodide 
solutions (1). 

2. Add carbon bisulfide to solutions of sodium chloride, potassium bromine, and 
potassium iodide. Shake. Is there a color change (2)? Now add a little chlorine 
water to each tube and again shake. In which tubes do you note a liberation of 
free halogens (3)? 


85. Hypochlorites. 

(a) The general preparation of a hypochlorite is by passing chlorine into a cold solution 
of an alkali. Give equation for the preparation of potassium hypochlorite (1). 

(b) Make a paste by mixing a little bleaching powder with water. Test its bleaching 
effect on colored calico (1). Why must moisture be present before bleaching will occur 
(2)? Add a little dilute sulfuric acid to the paste. What is the gas liberated (3)? 
Give equation (4). Is bleaching more rapid after addition of the acid (5)? Why (6)? 


86. Chlorates. 

(a) The general preparation of a chlorate is by passing chlorine into a hot solution of an 
alkali. Give equation for preparation of potassium chlorate (1). 

(b) Examine some crystals of potassium chlorate as to color, taste, and solubility (1). 
Recall the effect of heating potassium chlorate. Why is potassium chlorate (chlorate 
of potash) used as a gargle in cases of sore throat (2)? Why is it used in many ex¬ 
plosives (3) ? To a few crystals of potassium chlorate in a test tube add two or three 
drops of concentrated sulfuric acid. (Caution: Drop acid by means of a glass tube 
and do not look down the tube, as the mixture is very explosive.) What are the green¬ 
ish fumes (4)? Test them with a burning splinter (5). 


46 





















. 










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' 










CARBON AND CARBON DIOXIDE 


(FIFTH WEEK) 


87. Carbon. 

(a) Heat some sawdust in a test tube. Are the gases evolved combustible (1)? What are 
the drops of oily liquid which collect on the cool portions of the tube (2)? How does 
this liquid react to litmus (3)? What is the residue left in the tube (4)? 

(b) Allow a small luminous flame to play on the dry outside of a beaker of cold water. 
What is the deposit (1)? What is the name given to this form of carbon (2)? What 
other forms exist (3)? 

(c) In your evaporating dish burn a few drops of kerosene, holding a dry beaker of cold 
water in the smoky flame. What is the deposit (1)? Repeat, using turpentine. Which 
gives the more smoky flame (2)? Why (3)? 

(d) In separate test tubes heat small quantities of starch and sugar.. What vapors are 
driven off (1)? What is the residue (2)? In general, what classes of compounds give 
carbon when heated (3)? 

(e) Allow a direct flame to play upon a piece of charcoal. Does carbon bum (1)? Is 
carbon soluble in acids (2)? In alkalies (3)? If insoluble how would you remove a 
carbon deposit from an evaporating dish (4)? Why is oxygen used in removing car¬ 
bon from the cylinders of gas engines (5)? 


88. Carbon Dioxide. 

(a) Fit an Erlenmeyer flask with two-hole stopper, thistle tube, and delivery tube. In the 
flask place a few pieces of marble or pulverized limestone or calcium carbonate. Pour 
enough water in the flask to just cover the end of the thistle tube and then add 10-15 
cc. concentrated hydrochloric acid. Collect two bottles of the gas by downward 
displacement of air. Give equation (1). 

(b) Test one bottle with a burning splinter. Is the gas combustible (1)? Does it support 
combustion (2)? Pour a little limewater in the bottle and shake. What is the pre¬ 
cipitate (3)? Give equation (4). 

(c) Determine whether the gas is heavier or lighter than air by pouring the gas from the 
second bottle into an empty bottle. Test each bottle with limewater as before (1). 

(d) By means of a glass tube blow your breath through some limewater in a test tube. 
What gas is a constitutent of the air exhaled from the lungs (1)? 

(e) Hold the mouth of an empty bottle over a low flame. Add a little limewater and 
shake. The presence of what gas is shown (1)? 

(f) Add a little hydrochloric acid to small pieces of ammonium carbonate, potassium 
carbonate, and sodium carbonate in separate test tubes. Test the gas evolved. (A 
simple test is to suspend a drop of limewater on the end of a glass tube and hold 
it in the vapor given off from the tube. A milky appearance to the drop indicates the 
presence of carbon dioxide (1).) 

(g) In general, carbon dioxide is evolved when any carbon compound is burned in air, or 
by the action of any acid on any carbonate. Is the gas soluble in water (1)? How 
could you prepare carbonated water (2)? Is carbonic acid a weak or strong acid (3)? 
Stable or unstable (4)? 





























OPTIONAL EXPERIMENTS 
MIXTURES AND COMPOUNDS 

89. Two-Part Mixture. — Grind together in a mortar about 4 grams of sulfur and 6 grams 
of iron, pulverized. Divide into four portions. 

(a) On first portion try the effect of a magnet secured from stock room (1). What is 
removed from the mixture (2)? 

(b) Place second portion in a test tube and add 5 cc. of carbon bisulfide. Shake and 
filter, collecting the clear filtrate in a small watch glass and allow to evaporate in the 
air. (Carbon bisulfide and its fumes are very inflammable and should be kept away 
from a flame.) Examine the residue after evaporation (1). What was removed by the 
carbon bisulfide (2)? 

(c) Place third portion in a test tube and add a little dilute hydrochloric acid. Test the 
gas which escapes by means of a flame (1). Note odor (2). What part of the mixture 
seems to disappear (3)? 

(d) Heat the fourth portion of the mixture in a test tube until the mass begins to glow at 
the bottom. Remove tube from flame and note whether combustion continues (1). 
Is heat given off in this action (2)? When the action is over, strongly reheat the lower 
end of the tube until excess of sulfur has been driven off. Why was an excess of 
sulfur used (3)? Carefully break off the lower end of the tube by dropping a little 
cold water on it, while it is still hot. Remove the pieces of glass and grind the product 
in a mortar. Divide into three portions. 

Test one portion with a magnet (4). 

Test a second portion with carbon bisulfide as before. Is there a residue on the 
watch glass (5)? 

Treat the last portion with a little dilute hydrochloric acid. Note the odor. Is 
there a residue left undissolved (6)? 

Compare the properties of the mass after heating with its properties before heating 

(7). 

Has a chemical change been accomplished (8)? If so, what is the name of this 
product (9)? Define physical mixture, homogeneous mixture, chemical compound, 
chemical change, and element (10). 


90. In clean mortar rub together a globule of mercury and a pinch of powdered sulfur. Is 
the black mass a physical mixture or a chemical compound (1)? How proved (2)? To what gen¬ 
eral class of elements does mercury belong (3)? What one characteristic distinguishes mercury 
from all other elements of this class (4)? What is an amalgam (5)? Have you formed an amal¬ 
gam in this experiment (6)? 


48 









91. Analysis of Gunpowder — A Three-Part Mixture. — In a weighed evaporating dish weigh 
out 2 grams of black gunpowder. Transfer the powder to a filter paper in a f unn el and wash 
five times with sufficient hot water to cover the material, catching the filtrate in a clean beaker. 
Evaporate this aqueous solution in the weighed evaporating dish over boiling water to avoid any 
loss by spattering. 

When thoroughly dry, weigh dish and residue and record weight. What is the residue (1)? 

The residue on the filter in the funnel is dried thoroughly in an oven or over a very low flame 
and then washed five times with 10 cc. portions of carbon bisulfide. The filtrate from this 
washing is allowed to evaporate in the air or by blowing air over the surface of the liquid in the 
weighed evaporating dish or a watch glass. Record this weight. What is the residue (2)? Why 
is the residue dried thoroughly before washing with carbon bisulfide (3)? 

The residue on the paper after the carbon bisulfide washing is now dried on a glass over 
boiling water and weighed, using a counterposed filter paper. Record this weight. What is this 
residue (4)? 

Calculations: 


Wt. of residue from water solution X 100 
Wt. of original sample 


%'of KN0 3 . 


Wt. of residue from carbon bisulfide X 100 
Wt. of original sample 


% of S. 


Wt. of residue on filter paper X 100 ^ ^ 

Wt. of original sample 

Write the equation, showing the chemical reaction taking place when gunpowder is exploded (5). 
Calculate the volume of COa at laboratory temperature and pressure which is liberated if the 
2-gram sample of gunpowder is burned (6). 

Under similar conditions what volume of nitrogen is produced (7)? 


49 












92. Determination of Water of Crystallization. (Two Students.) (a) Weigh accurately on 
analytical balance a porcelain crucible, then put into the crucible about 2 grams of crystallized 
barium chloride, and weigh carefully again. The difference in the weights represents the actual 
weight of barium chloride. Support the crucible in a pipe-stem triangle so adjusted in height 
that the bottom of the crucible is a short distance above the top of the inner cone of the Bunsen 
flame. Heat the crucible very gently at first, then increase the heat to full intensity of the flame 
for about fifteen minutes. Slowly cool the crucible and weigh the contents. Heat again, weigh 
until a constant weight is obtained. Record this weight. 

From your data ascertain: (1) the weight of the crystallized barium chloride; (2) the weight 
of the anhydrous barium chloride after heating. 

The “anhydrous” salt is the crystallized salt minus its water of crystallization. The weight 
in grams of these two substances are in the same ratio as their molecular weights. 

Calculate: (1) the molecular weight of the anhydrous barium chloride; (2) number of mole¬ 
cules of water in crystallized barium chloride; (3) percentage of water in the compound. 

(b) Repeat the experiment with copper sulfate. 

TABLE OF DATA 

(a) (b) 

Weight of cruc. + barium chloride. = 

Weight of cruc. empty. = 

Weight of cryst. barium chloride. = 

Weight of cruc. + anhyd. barium chloride after 

heating. = 

Weight cruc. empty. = 

Weight anhyd. barium chloride. = 

Mol. weight of crystallized barium chloride. = 

Mol. weight of anhydrous barium chloride. = 

Mol. weight of water in crystallized barium chloride = 

Formula of crystallized barium chloride. = 


50 
























93. (Four Students.) Equivalent of Sodium and Percentage of Hydrogen in Water. — 
Apparatus: A piece of glass tubing about two or three centimeters in length sealed at one end in 
the flame, analytical balance and weights, 1000 cc. graduated cylinder, 50 cc. burette with pinch 
clamp, and small wad of filter paper to be used to loosely plug the glass capsule. 

Procedure: 

(a) Weigh the glass capsule and paper plug and record weight. Dry a piece of metallic 
sodium on a clean, dry filter paper, cut a fresh surface on the metal, and press the 
open end of the capsule into the bright metal until about a centimeter length of the 
sodium is forced in the tube. (Caution: The action of sodium and potassium is 
very violent in water, and the hands and the metal must be kept perfectly dry.) 
Now loosely plug the capsule with the paper wad, weigh, and record the weight. 

Fill the burette with water, invert in the cylinder of water, see that the level of 
water in the burette is the same as that in the cylinder, and record the exact reading 
of the water in the burette. Now with the plugged end down, insert the capsule con¬ 
taining the sodium in the burette. 

When the action has ceased, raise or lower the burette until the water levels are 
again the same and record the exact reading of the water in the burette. Record the 
temperature of the water in the cylinder, and the barometric reading. Correct the 
volume of hydrogen for aqueous pressure, reduce to standard conditions, and cal¬ 
culate the weight of sodium needed to liberate 1 gram of hydrogen. What term is 
applied to this weight? 

(b) Repeat this experiment, using metallic potassium. 

TABLE OF DATA 

(a) (b) 

Weight of capsule + sodium + plug. = 

Weight of capsule + plug. = 

Weight of sodium used. = 

First burette reading. = 

Burette reading after evolution. = 

Volume of hydrogen evolved. = 

Barometric pressure. = 

Temperature of water. = 

Aqueous pressure at this temperature.= 

Correct pressure of hydrogen. = 

Volume of hydrogen at S. C. = 

Weight of hydrogen. = 

Equivalent or combining weight of sodium. = 

(c) In (a) and (b) calculate the weight of: 

(1) Water required to react with the metal. = 

(2) NaOH or KOH produced. = 

(3) Hydrogen in 100 grams of water. = 

(d) Calculate percentage of hydrogen in H 2 0. = 

Correct percentage of hydrogen in H 2 0. = 

Percentage of error. = 

Possible sources of error. = 


51 

































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94. Analysis of Fabric Containing Wool, Silk, and Cotton. — A weighed portion of the sample 
is treated for ten minutes with a cold solution of nickel hydrate in ammonia. This will dissolve 
any silk present. Wash well, dry, and reweigh. The loss in weight represents silk. The residue 
is next boiled for ten minutes in a 5 per cent solution of caustic potash. This will dissolve any 
wool present. Wash well, dry, and reweigh. The loss in weight represents wool, while the residue 
consists of cotton. Report percentage of each present in fabric. 


95. Distinction Between True Silk and Artificial Silk. — Artificial silk (or lustra-cellulose) is 
a fiber prepared from a solution of collodion or other cellulose solution. It consists of cellulose, 
whereas true silk is a nitrogenous animal substance. Artificial silk burns readily in the air like 
cotton, without evolving a strong odor, while silk burns slowly. A weighed portion of the sam¬ 
ple is treated at ordinary temperature for twenty minutes with an alkaline solution of copper 
sulfate. This will completely dissolve the natural silk, leaving the artificial fiber as a residue. 
The latter is thoroughly washed, dried, and reweighed. The alkaline solution of copper sulfate 
is prepared by dissolving 10 grams of copper sulfate in 100 cc. of water and 5 cc. of glycerin; a 
strong solution of caustic soda is then added until the precipitate at first formed just redissolves. 


96. To Distinguish Between Cotton and Linen. 

(a) Steep the sample containing these two fibers for two minutes in concentrated sulfuric 
acid; wash well with water, gently rub with the fingers, and finally steep in dilute 
ammonia; then squeeze and dry. The cotton fiber will be converted into a jelly-like 
mass by the action of the acid, and is more or less completely removed by the rubbing 
and washing. The linen remains but little altered. By weighing the sample before 
and after the treatment an approximate idea of the amounts of cotton and linen pres¬ 
ent may be obtained. 

(b) Steep the sample to be tested in olive oil; then press between filter paper to remove 
the excess of oil. The linen fibers will become gelatinous in appearance and translu¬ 
cent, whereas the cotton remains unaltered. When placed on a dark background the 
linen fibers will now appear dark and the cotton fibers light. 

(c) Steep the sample to be tested in an alcoholic solution of rosolic acid, and then in a 
strong solution of caustic soda; finally rinse in water. The linen fibers will become 
rose-colored while the cotton is colored much lighter and most of the color is removed 
by the rinsing. None of these tests are very satisfactory when the linen has been 
bleached, for then its cellulose is practically identical with that of cotton. The most 
satisfactory means of qualitatively distinguishing linen from cotton is by a micro¬ 
scopic examination, as these fibers exhibit very different microscopic properties. 


52 






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ANALYSIS OF SOAP 


97. 


(a) Insoluble Matter. — Dissolve 5 gm. of the sample in 75 cc. of water, heating if 
necessary. If there are more than mere traces of residue, filter on a tared filter paper, 
wash with hot water until the filtrate is neutral, dry at 150° C., and calculate the 
per cent. 

Test various “hand” and scouring soaps for insoluble matter. 

Alkali may exist in soap in at least three forms — free, combined, and as alkaline 
carbonates, borates, etc. 

(b) Detection of Free Alkali. — Treat the freshly cut surface of the soap with a few drops 
of the alcoholic solution of phenolphthalein. If no red color appears, it may be as¬ 
sumed that free alkali is absent. Take great care that no water comes in contact 
with the soap, otherwise the results may be misleading. 

When testing washing powders for free alkali, dissolve a small quantity in alcohol 
and add the phenolphthalein. 

(c) Combined Alkali. — If free alkali, alkaline carbonates, borates, etc., are absent, 
dissolve 1 gram of the soap in 20 cc. of hot water. Treat with phenolphthalein. A 
red color is immediately seen. Why? 

Titrate with N/10 oxalic acid. Allow the acid to drop slowly from the burette until 
the last drop added destroys the pink color. Shake or stir the solution frequently 
during the operation and be sure to waste none. If after a minute the pink color 
reappears, add another drop of the acid. 

Free alkali may also be estimated by dissolving it from 1 gm. of the soap by means 
of alcohol. Filter, wash with alcohol, and titrate as for combined alkali. Calculate 
as NaOH. 

Calculation for Free or Combined Alkali. — One cc. of N/lO oxalic acid neutralizes 1 
cc. of N/lO NaOH, or 0.004 gm. of NaOH. Suppose the number of cubic centimeters 
of acid required is 4.75; then the per cent of alkali present is 4.75 X .004 = .019 
or 1.9 per cent. 

Write the reaction between oxalic acid and sodium hydroxide. 

(d) Total Alkali. — If the soap is free from sand and other mineral matter, burn 2 gm. to 
ash in a porcelain crucible. Cool and wash the contents into 50 cc. of water. Boil 
and titrate with N/lO oxalic acid, using two drops of methyl orange as an indicator. 
Express the result as the acid number; that is, the number of cubic centimeters of 
decinormal acid necessary to neutralize the alkali in 1 gm. of the soap. 

(e) Alkaline Carbonates. — Place 5 gm. of the soap in an Erlenmeyer flask and add 20 cc. 
of alcohol. Set a funnel in the neck of the flask to act as a reflux condenser, and heat 
on a water bath for ten minutes. Alkaline carbonates will be in the residue. Place 
a bit of this residue on a clean platinum wire and apply the flame test for sodium and 
potassium. Filter the alcohol and dissolve the residue in warm water. Add a few 
drops of dilute HC1. A marked effervescence indicates the presence of carbonates; 
whether Na 2 CC >3 (washing soda) or K 2 C0 3 (potash) will be indicated by the flame test. 

(f) Borates. — Place a strip of turmeric paper in a watch glass. Add a few drops of the 
soap dissolved in water, and acidify slightly with dilute HC1. Evaporate to dryness 
over a water bath. The presence of borates is indicated by the pronounced reddening 
of the dried paper. 


53 


/ 







IDENTIFICATION OF VEGETABLE OILS 


98. Halphen Test for Cottonseed Oil. — Mix CS-» containing about 1% of sulfur in solution 
with an equal volume of amyl alcohol. Mix equal values of this reagent and the oil sample and 
heat in a bath of boiling, saturated brine for one to two hours. In the presence of as little as 1% 
of cottonseed oil, a characteristic red or orange-red color is produced. 

Lard and lard oil from animals fed on cottonseed meal will give a faint reaction; their fatty 
acids also give this reaction. 

The depth of color is proportional, to a certain extent, to the amount of cottonseed oil present, 
and by making comparative tests with cottonseed oil some idea as to the amount present can be 
obtained. Different oils react with different intensities, and oils which have been heated from 
200-210° C. react with greatly diminishing intensity. Heating for ten minutes at 250° C. renders 
cottonseed oil incapable of giving the reaction. 

Note. — 

(1) Brown cottonseed oil and old rancid oil cannot be identified by this test. 

(2) A blank test should always be conducted under the same conditions on a pure sample 
of the oil being tested and also on a pure oil to which has been added a little cottonseed 
oil. 


99. Bellier Test for Peanut Oil. — Place 1 gram of the sample in a long test tube. Add 
5 cc. of alcoholic KOH solution. Boil gently over a small flame, holding the burner in the hand until 
saponification is complete, as shown by homogeneous solution (generally three to five minutes). 
Add the proper amount of acetic acid (see below) to exactly neutralize the 5 cc. of alcoholic KOH. 
Mix well, cool rapidly in water at about 17° C., and let stand in the water for at least thirty minutes, 
shaking occasionally. Then add 50 cc. of 70% alcohol containing 1% by volume of con. HC1 and 
again place in the water for one hour. If no peanut oil is present, a clear, opalescent liquid is 
formed. If more than 10% of peanut oil is present, a flocculent, crystalline precipitate remains. 
Even with 5% of peanut oil a distinct precipitate remains and separates on standing. 

(1) Alcoholic KOH. Dissolve 8.5 grams of pure KOH in 70% alcohol, and dilute to 
100 cc. with the alcohol. 

(2) Acetic acid: This should be of such strength that 1.5 cc. will exactly neutralize 5 cc. 
of the above solution. The dilute acetic acid reagent (4.10) is approximately the cor¬ 
rect strength but should be tested against the alcoholic KOH. 

Elaidin Test. Place in a test tube 10 grams of the oil, 5 grams of con. HN0 3 , and 1 gram of 
mercury, and dissolve the latter by shaking continuously three minutes. Let the mixture stand 
for twenty minutes and again shake for one minute. The behavior of different oils after that time 
is recorded in the following table: 


Kind of Oil 

Olive 

Arachis (peanut) 
Colza (rape) 
Linseed 


Consistence 

Solidified after 60 minutes. 
Solidified after 80 minutes. 
Solidified after 185 minutes. 
Forms red doughlike scum. 


Notes. — 

(1) The test must be made at a temperature not lower than 25° C. and the temperature 
must be uniform throughout the experiment. 

(2) The length of time required for solidification is of far greater importance than the 
ultimate consistency of elaidin formed. 

(3) The test cannot be made qualitative. Also the age of the oil and the length of time 
it has been exposed to air and fight have an important bearing on the results. It is 
necessary to carry out the test side by side with an oil of known purity under exactly 
similar conditions. 


54 






















100. Maumene Test of Oils. — Weigh accurately 50 grams of the oil into a 250 cc. beaker. 
Have ready a bottle of con. H 2 SO 4 , the exact strength of which has been determined by titration. 
Place the bottle of acid and the beaker of oil in a large vessel of water until both have acquired 
the same temperature, which should be about 20° C. Remove the beaker of oil, wipe the outside, 
and place in a “nest’' of cardboard having hollow sides stuffed with cotton wool, or in a large 
beaker lined with cotton wadding. Immerse the bulb of a centigrade thermometer in the oil 
and note the temperature. Add 10 cc. of con. H2SO4, by means of a pipette, and let it run rapidly 
into the oil. The time allowed for emptying of the pipette should be only one minute. During 
this time, stir the oil with the thermometer and continue stirring until no further rise of temper¬ 
ature is observed. The highest point is easily noticed, as the temperature remains constant for 
some little time before it begins to fall. 

Note. — Olive oil is characterized by low Maumene and iodine values and by solid Elaidin. 


101. Babcock Test. Determination of Butter Fat in Milk. — Before proceeding with the 
determination, the following apparatus must be prepared and adjusted; and the apparatus 
calibrated: 

A. Apparatus: Centrifuge. 

B. Six-inch milk test bottles, on supply at the stock room, 8 % milk test bottles, 6 inch 

size. The bottle holds 45 cc. but extra space is provided for the water and acid. 

C. Pipettes; graded to determine 17.6 cc. water. 

D. Graduates: capacity 17.6 cc. The 25 cc. graduates are satisfactory; such as provided 
in stock room. 

Calibration of Apparatus: Be sure that each space on the neck of the bottle measures 
1 cc. Test by using water and reading at the bottom of the meniscus line. 

The limit of error should not be more than .5 per cent. 

Material for Testing: Obtain from the instructor. Milk has been provided by the 
Dairy Husbandry Department. 

Pipette 17.6 cc. of the carefully mixed sample into a test bottle. Add 17.5 cc. of the concen¬ 
trated H2SO4 solution (Sp. gr. 1.82-1.83). Mix and when the curd is distinct, centrifugalize for 
4 minutes at the required speed for the machine used. Add boiling water to the neck of the bottle, 
and revolve for one minute. Again add boiling water so as to bring the fat within the scale on the 
neck of the bottle, and after whirling for one minute more, read the length of the fat column, 
making the reading at 56-60° C., at which temperature the fat is wholly liquid. The reading gives 
directly the per cent of fat in the milk. 


102. Determination of Arsenic (As 2 0 3 ) in Dipping Vat Mixture. — Place 25 cc. of the dip in 

a large beaker. Dilute with 200 cc. of distilled water, acidify with HC1, and add sodium bicarbonate 
(NaHC0 3 ) until slightly alkaline. To this solution add 5 cc. of starch solution and then standard 
iodine solution from a burette until the solution becomes a blue color, stirring constantly. From 
these results figure the percentage of arsenic in the dip, considering the weight of 25 cc. of dip to 
be 25 grams. 

Note. — The dip is made according to the following proportions: white arsenic 8 lb., sodium 
carbonate 24 lb., made up to 500 gal. with water. 


55 





103. Determination of Arsenic in Dusting Powder or Wall Paper. — Weigh on the balance 
2 grams of calcium arsenite or cotton dusting powder. Dissolve in 25 cc. of HC1 solution (12.5 
cc. of con. HC1 and 12.5 cc. of water), dilute to 200 cc. with distilled water. To 25 cc. of this 
solution, in a 400 cc. beaker add 200 cc. of distilled water, sodium bicarbonate in excess, 5 cc. of 
starch solution and standard iodine solution from a burette until a blue color is obtained. From 
the results figure the percentage of arsenic (As 2 0 3 ) in the sample. 

Note. — If arsenic is present as calcium arsenate it must first be reduced to arsenite. 


104. Determination of Soluble Copper in Bordeaux Mixture. — Weigh on a rough balance 
2 grams of copper sulfate and dissolve in 50 cc. of water. Add calcium hydroxide solution until 
there is no soluble copper in the solution. To test for soluble copper filter about 5 cc. of the 
solution and add potassium ferrocyanide solution. Should you obtain a brick red precipitate it 
is an indication of soluble copper present in the solution. After you have added enough lime 
to the solution so that you cannot obtain a test for soluble copper, then pass carbon dioxide into 
the solution until you again obtain a test for soluble copper, testing in the same way as you did 
before. Explain the chemistry of the reaction and give reason for each result obtained. 


105. Determination of Lead in White Lead. — Place in a 400 cc. beaker about 1 gram of 
pure white lead. Cover the powder with 25 cc. of nitric acid (1-3), heat gently until complete 
solution takes place, and then dilute with 150 cc. of distilled water. Add 10 cc. of sulfuric acid 
(1-3). Evaporate on wire gauze over very low flame until the odor of nitric acid can no longer be 
detected and until the dense white fumes of sulfuric acid begin to appear. Cool, add 50 cc. of 
distilled water and 50 cc. of 95% alcohol. Set aside in a cool place for three or four hours and 
filter through a weighed Gooch crucible using suction flask. Wash three times with a (1-3) 
solution of alcohol and dry in oven for one hour at temperature of 180°-200° C. Weigh and 
calculate the percentage of lead in the original sample. 

In white lead paint how would you remove the linseed oil before proceeding with this experi¬ 
ment? 

Give equations for the commercial preparation of white lead. 


56 


k 







URINANALYSIS 


106. Preparation of Sample. — Before making the tests the urine must be prepared, so as to 
remove as many of the interfering substances as possible. Boil 10 cc. of the urine, add 5 cc. hot 
lead acetate, mix and filter, remove excess of lead from the filtrate by adding 5 cc. Na 2 HP0 4 and 
filter again. Use a little infusorial earth if necessary to obtain a clear filtrate. 

(a) Detection of Glucose. 

Fehling’s Test. Boil 6 cc. Fehling’s Solution for several minutes and if no pre¬ 
cipitate forms add three to four drops of prepared urine at a time until about 3 cc. 
have been added, boiling two minutes after each addition. To detect red Cu 2 0 which 
forms if dextrose is present, allow the test tube to stand so the oxide separates out 
and examine with reflected daylight. Small quantities are exceedingly hard to see 
by artificial light. 


(b) Detection of Albumin. 

Heat Test. Place some of the urine in a test tube and boil the upper portion over a 
small flame; the lower cold portion will show by contrast if the boiled portion has 
become more or less turbid, due to coagulation of the albumin. (Earthy phosphates 
and resin acids may cause a turbidity in this test, the former being soluble in acetic 
acid, the later insoluble in acetic acid, but soluble upon the addition of alcohol.) 
Heller’s Test. Place 5 cc. of con. HN0 3 in a test tube and float about the same 
quantity of urine upon it so as to form a distinct layer; albumen causes a turbid 
zone at the juncture of the two liquids. 

(c) Detection of Uric Acid. 

HC1 or HN0 3 causes a precipitation of uric acid from urates; in Heller’s albumen 
test this is frequently noticed in concentrated urines; the precipitate, however, does 
not form at the line of juncture, but some distance above the layer of urine. 
Murexide Test. Add a few drops of nitric acid to a small quantity of urine and 
carefully evaporate to dryness over a small flame; exposing the residue to the vapors 
of NH 3 produces a purple color due to the formation of murexide or ammonium pur- 
purate. 

Ammonium Chloride Test. Add solid ammonium chloride to saturation and allow 
to stand for two hours. Uric acid precipitates as acid ammonium urate. 


57 










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■ ; li ~J . '• - ■ 







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J 






107. Silver Plating. — Solutions. Silver cyanide plating solution (very poisonous). 

Dissolve 396.2 grams of AgN0 3 in 4.8 liters of pure distilled water. Add a moderately con¬ 
centrated solution of NaCN until precipitation is complete. Avoid any excess of NaCN. Filter, 
wash precipitate, and dissolve the precipitate of AgCN in excess of NaCN solution. As soon as 
dissolved, dilute to 10 liters and boil for one hour, replacing water lost by evaporation by distilled 
water. Filter and throw residue into Ag waste jar. (Solution prepared by instructor.) 

Use an E. M. F. of 0.75 volt at 10 cm. distance with a current density of 0.3 amp. 

The bath must be agitated during the plating process. 

Preparatory solutions: 

No. 1. Hot 1 to 8 solution of KOH. 

No. 2. Pickle 1 to 10 H 2 S0 4 (for rough castings only). 

No. 3. Preliminary pickle: 

200 parts by weight con. HNO 3 . 

1 part by weight NaCl. 

2 parts by weight lampblack. 

No. 4. Bright dip: 

65 parts by weight HNO 3 . 

100 parts by weight con. H 2 S0 4 . 

1 part by weight NaCl. 

No. 5. Quickening or amalgamating: 

1.091 gms. Hg. 

965 cc. distilled water. 

Con. HN0 3 to clear solution. 

Procedure in Silver Plating. 

1. Select some small article that you wish to plate such as silver pencil, belt buckle, etc. 

2. Carefully smooth and polish surface with fine emery paper taking extreme care to remove 
all traces of lacquer from article. Finish polishing with jeweler’s rouge. (Article must be fully 
prepared as above before you come to class.) 

3. Attach small wire to article and immerse in solution No. 1 to remove all traces of grease and 
dirt. Caution: Do not touch article with the fingers after it has been in the solution No. 1. 
After rinsing in boiling H 2 0 dry article with the whiting. 

4. Rub well with cloth (ordinary laboratory towel will not do) dipped in pumice or jeweler’s 
rouge. (This cloth must be clean cotton or linen that has been laundered at least twice.) 

5. Dip in solution No. 3 until told to remove by instructor. 

6. Remove and rinse well in boiling H 2 0 and dry with clean cloth. DO NOT touch article 
with the fingers. 

7. Dip in solution No. 4 for an instant. Do not leave article in this solution for any length 
of time as it will become brittle. 

8. Thoroughly rinse article in several baths of boiling H 2 0. Let dry in the air and dip into 
solution No. 5. 

9. As soon as article gets white remove, rinse in hot H 2 0, and plunge into the electroplating 
bath. 

The following data must be taken during the experiment: Wt. of article to be plated as brought 
to the lab. Wt. of plated article as finished. Reading of voltmeter. Time at beginning of plating. 
Reading of ammeter. Time at end of plating. Weight of silver before using. Total time required. 
Calculate amount of silver added to article. Percentage of silver in plated article. Calculate 
in coulombs the amount of electricity consumed in plating process. Calculate amount of elec¬ 
tricity required to deposit 1 gram of silver. Write up this experiment in your notebook and 
draw a diagram of apparatus used. 


58 



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, 




' 


PART II 


ELEMENTARY QUALITATIVE ANALYSIS 





GENERAL DEFINITIONS AND INTRODUCTORY PRINCIPLES 


The branch of chemistry which concerns itself with the composition of substances is 
called Analytical Chemistry. Before an analysis is made to determine the amount of each 
substance present, it is necessary to discover what substances must be determined. Qualita¬ 
tive Analysis, the study of what substances are present therefore necessarily precedes 
Quantitative Analysis, which considers the amount of each substance found to be present. 

In our elementary qualitative analysis we will merely determine the metals and the 
acid radicals present and then report probable combinations. In order to correctly identify 
metals and acid radicals, it will be necessary for the student to run preliminary tests. These 
“ test-tube ” tests are extremely important, since it is only by associating similar results with 
these that we are able to identify the various metals and acid radicals in our “unknowns.” 
The student should perform each of these tests carefully, using small quantities of solution 
to be tested, adding reagent drop by drop, shaking between drops, and noting the effect of 
addition of small quantity and of excess of reagent. 

The principle most commonly used in qualitative analysis , is to add a reagent to the 
solution to be tested which will throw down a precipitate which can be recognized by its 
characteristic color, texture, and solubility in various other reagents. Colors are described 
as white, orange, canary yellow, black, brown, purple, muddy colored, slate, green, blue, etc. 
(Caution: The student should be careful in describing colors, to specify whether the solution 
or the precipitate is colored. This can be determined by allowing the precipitate to settle.) 

Texture of a precipitate may be described as: curdy, spongy, flocculent, gelatinous, 
finely divided, crystalline, etc. 

Solubility of a precipitate is best determined by collecting a small amount of the solid 
on a filter paper folded to fit a funnel in the wooden funnel stand, transferring a portion of 
this to a test tube, adding the solvent desired. Precipitates may be removed from the filter 
either by means of a spatula or by washing them with a small stream of water through a 
puncture made in the point of the filter. In testing whether a precipitate is soluble in excess 
of reagent, this may be accomplished in the same tube or beaker in which precipitation 
occurred. A table of solubility may be found at the back of this manual. 

■ A reagent is a substance used to bring about a chemical change. This may be a general 
reagent which acts similarly on a group of substances, or a specific reagent which has a 
characteristic action on only one substance. When a liquid containing a solid is filtered, 
the solid precipitate on the filter paper is termed the residue. The clear liquid running 
through the paper is the filtrate. It is always recommended to test the filtrate with a small 
amount of the reagent used to insure complete precipitation. 

A residue on a filter may be washed by pouring water over this residue several times, 
allowing it to filter through the paper. Frequently a granular precipitate is washed by 
decantation. In this process water is added to the precipitate in a beaker, the mixture 

61 


stirred, allowed to settle, and then the clear liquid above the precipitate poured off, and the 
process repeated. Liquids may be poured without spilling by laying a glass rod over the 
lip of the beaker and pouring down the glass rod into the filter. In case of a gelatinous 
precipitate which clogs the pores of the paper, decantation is recommended. Under certain 
conditions a finely divided precipitate may pass through a filter paper. In such cases it is 
advisable to heat the mixture to enlarge the particles and, if necessary, run through double 
filter papers. In case of difficult filtration, heating is recommended as above. Filtration 
may be speeded by attaching a piece of glass tubing of small diameter to the funnel tube, 
making the total length ten or twelve inches. 

In qualitative analysis it is absolutely necessary to avoid contamination of reagents. 
Stoppers of reagent bottles must not be laid on desk or placed in another bottle. Glassware 
used must be washed clean in tap water and rinsed with distilled water. In making up 
solutions or in dilutions, only distilled water is used. All chemicals used are chemically 
pure (C. P.) and must be kept pure. To avoid mistakes or confusion, if a residue or filtrate 
is set aside for future use, it should be put in a clean, stoppered Elrenmeyer flask and labeled 
in such a way that the student may know what it is. 


THE GROUP OF METALS 

For the purpose of qualitative analysis it has been found convenient to divide the metals into 
six groups. The members belonging to each group together with the group precipitant are as 
follows: 

Group 1. Lead, Silver, Mercury (ous). 

Group Reagent, HOI. 

Group 2. Subdivision A. Mercury(ic), Lead, Bismuth, Copper, Cadmium. Sulfides are in¬ 
soluble in ammonium polysulfide. 

Subdivision B. Arsenic, Antimony, Tin. Sulfides are soluble in ammonium poly- 
sulfide. 

Group Reagent. Hydrogen Sulfide with HC1 solution. 

Group 3. Iron, Aluminum, Chromium. 

Group Reagent. Ammonium hydroxide in the presence of ammonium chloride. 

Group 4. Cobalt, Nickel, Manganese, Zinc. 

Group Reagent. Ammonium sulfide. 

Group 5. Barium, Strontium, Calcium. 

Group Reagent. Ammonium Carbonate. 

Group 6. Magnesium, Sodium, Potassium, Ammonium. “Soluble” group. 


62 


GENERAL DIRECTIONS 

When making the test reactions always use a small quantity of the given solution 
about 5 cc. is enough. Add the reagent slowly, a few drops at a time, shaking the test tube 
after each addition to thoroughly mix the solutions. Note carefully the first appearance of 
a precipitate, or other evidence of reaction, and any changes on the addition of more of the 
reagent. Finally add an excess of the reagent and note its effect. 

When a precipitate forms, its color and character (crystalline, curdy, gelatinous, floccu- 
lent, etc.) should be noted and recorded. If no reaction is observed, this fact should also 
be recorded. Keep the record carefully, as it will be of great value in the subsequent work, 
especially in verifying the analysis of unknown solutions. 

On blank page opposite write equations for all reactions underlining precipitates, stat¬ 
ing color and solubility where called for, and indicate gases evolved by arrow up. When 
precipitate dissolves in some reagent, write reaction. 


63 




EXERCISE 1. —LEAD 

Before making the following experiments read caref ully the General Directions on preceding page. 
1. Try the action of each of the following reagents on lead acetate or lead nitrate solution: 

(a) Hydrochloric acid. Is ppt. soluble in boiling H 2 0? Is ppt. soluble in NH 4 OH? 

(b) Potassium chromate. Test the solubility of the precipitate in NaOH. 

(c) Potassium chromate. Is the precipitate soluble in acetic acid? 

(d) Dilute sulfuric acid. 

(e) Hydrogen sulfide. Test the solubility of the precipitate in dilute HC1. 

(f) Potassium iodide. 

(g) Ammonium hydroxide. 


64 


Name 


Date 


Section No. 


LEAD SALTS 

(a) (CH 3 COO) 2 Pb + HC1 = 

PbCl 2 + hot H 2 0 = 

PbCl 2 + NH 4 OH = 

(b) (CH 3 COO) 2 Pb + K 2 Cr0 4 = 

PbCr0 4 + NaOH = 

(c) PbCr0 4 + CH3COOH = 

(d) (CH 3 COO) 2 Pb + H 2 S0 4 = 

(e) (CH 3 COO) 2 Pb + H 2 S = 

PbS + HC1 = 

(f) (CH 3 COO) 2 Pb + KI = 

(g) (CH 3 COO) 2 Pb + NH 4 OH = 

(a) Pb(N0 3 ) 2 + HC1 = 

(b) Pb(N0 3 ) 2 + K 2 Cr0 4 = 

(d) Pb(N0 3 ) 2 + H 2 S0 4 = 

(e) Pb(N0 3 ) 2 + H 2 S = 

(f) Pb(N0 3 ) 2 + KI = 

(g) Pb(N0 3 ) 2 + NH 4 OH = 


65 




















































» 








- 































EXERCISE 2. —SILVER 


1. Try the action of each of the following reagents on silver nitrate solution: 

(a) Hydrochloric acid. Test the solubility of the precipitate in dilute HNO3. 
soluble in NH 4 OH? 

Add a little water and boil. Is the ppt. soluble in boiling water? 

(b) Potassium chromate. Test the solubility of the precipitate in NH4OH. 
soluble in NaOH? 

(c) Dilute sulfuric acid. 

(d) Hydrogen sulfide. Test the solubility of the precipitate in dilute HNO3. 

(e) Potassium iodide. Is the precipitate soluble in NH 4 OH? 


Is the ppt. 


Is the ppt. 


68 


Name 


Date 


Section No. 


SILVER SALTS 


(a) AgNOa + HC 1 = 

AgCl + HNO. = 

AgCl + hot H 2 0 = 

(b) AgNOa + K2CTO4 = 
Ag2Cr0 4 + NH4OH = 

(c) AgN 0 3 + H2SO4 = 

(d) AgNOa + H 2 S = 

Ag 2 S + HN 0 3 = 

(e) AgNOa + KI = 

Agl + NH4OH = 


69 
















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EXERCISE 3. — MERCURY(OUS) 

1. Try the action of each of the following reagents on mercurous nitrate solution: 

(a) Hydrochloric acid. Is the ppt. soluble in boiling H 2 0? 

Is the ppt. soluble in NH 4 OH? 

(b) Potassium chromate. Test the solubility of the precipitate in NaOH. 

(c) Dilute sulfuric acid. 

(d) Hydrogen sulfide. 

(e) Potassium iodide. 


72 


Name, 


Date, 


Section No. 


MERCUROUS SALTS 


(a) HgNOs + HC1 = 
HgCl + NH 4 OH = 
HgCl + hot H 2 0 = 

(b) HgN0 3 + K 2 Cr0 4 = 
Hg 2 Cr0 4 + NaOH = 

(c) HgNOs + H 2 S0 4 = 

(d) HgNOs + H 2 S = 

(e) HgNOs + KI = 


73 




























» 















EXERCISE 4. — MERCURY(IC) 

1. Try the action of each of the following reagents on mercuric chloride solution: 

(a) Hydrogen sulfide. Test the solubility of the precipitate in dilute HC1. In 50% 

hno 3 . 

(b) Dilute sulfuric acid. 

(c) Potassium iodide. 

(d) Potassium chromate. 

(e) Stannous chloride. 


76 


Name 


Date 


Section No. 


MERCURIC SALTS 

(a) HgCl 2 + H 2 S - 
HgS + HC1 = 

HgS + HNO s = 

(b) HgCl 2 + H 2 S0 4 = 

(c) HgCl 2 + KI = 

(d) HgCl 2 + K 2 Cr0 4 = 

(e) HgS + Aqua Regia (3HC1 + HNO 3 ) = 

(f) HgCl 2 + SnCl 2 = 


77 







EXERCISE 5.— BISMUTH 


1. Try the action of each of the following reagents on the solution of bismuth nitrate 
bismuth chloride: 

(a) Hydrogen sulfide. Is the precipitate soluble in dilute HC1? 

Is the ppt. soluble in 50% HNO 3 ? 

(b) Dilute sulfuric acid. 

(c) Ammonium hydroxide. 

(d) Potassium iodide. 

(e) Potassium chromate. 

(f) Water — Is the precipitate soluble in HC1? 


or 


80 


Name 


Date 


Section No. 


BISMUTH SALTS 


(a) Bi(N0 3 )3 + H 2 S = 

BLS3 + HC 1 - 
Bi 2 S 3 + HNO3 = 

(b) Bi(NOs)3 -t- H 2 S0 4 = 

(c) Bi(N0 3 )3 4* NH4OH = 

BiOOH + Na 2 Sn0 2 * = 

*(SnCl 2 + NaOH until ppt. just dissolves.) 


(d) 

Bi(N0 3 )s 

. + KI = 

(e) 

Bi(N0 3 )j 

, + K 2 Cr0 4 

(f) 

Bi(N0 3 )s 

» + h 2 0 = 


Bi(OH) 3 

+ HC1 = 

(a) 

BiCls + 

H 2 S = 

(b) 

BiCla + 

H 2 S0 4 = 

(c) 

BiCla + 

nh 4 oh = 

(d) 

BiCls + 

KI - 

(e) 

BiCL + 

K 2 Cr0 4 = 

(0 

BiCl 3 4~ 

H 2 0 = 


81 










EXERCISE 6.—COPPER 

1. Try the action of each of the following reagents on a solution of copper sulfate: 

(a) Hydrogen sulfide. Try to dissolve the precipitate by boding with dilute sulfuric acid. 
Is ppt. soluble in 50% NH0 3 ? 

(b) Dilute sulfuric acid. 

(c) Ammonium hydroxide. Add a drop at a time untd an excess is present. Now try the 
effect of H 2 S. 

(d) Make alkaline with NH 4 OH (test with litmus) and add KCN solution till blue color 
disappears. Now add (NH 4 ) 2 S. Explain. 

(e) Potassium iodide. The precipitate is cuprous iodide. Notice the color of the solution. 
Test it with starch paper. 

(f) Potassium chromate. 

(g) Potassium ferrocyanide. 


84 


Name 


Date 


Section No. 


COPPER SALTS 


(a) C11SO4 -f- H2S — 

CuS H2SO4 = 

CuS + HNOs = 

(b) CuSOi + H 2 S 0 4 = 

(c) CUSO4 + NH4OH = 
Cu(NH 3 ) 4 S 0 4 + h 2 s = 

(d) CuSO* + NH 4 OH + KCN = 

(e) C11SO4 + ia - 

(f) CuS 0 4 + K 2 Cr 0 4 = 

(g) CuS 0 4 + K4Fe(CN)e = 
Cu(N 0 3 ) 2 + H 2 S = 

Cu(N 0 3 ) 2 + NH4OH - 
Cu(N 0 3 ) 2 + NH4OH -f KCN = 
Cu(CN) 2 + (NH 4 ) 2 S = 


85 






















































































































































EXERCISE 7.— CADMIUM 


1. Try the action of each of the following reagents on a solution of cadmium nitrate, or cad- 
mium sulfate: 

(a) Hydrogen sulfide. Try to dissolve the precipitate in boiling dilute II 2 SO 4 . Is ppt. 
soluble in 50% HN0 3 ? 

(b) Sulfuric acid. 

(c) Ammonium hydroxide. 

(d) Make alkaline with NH 4 OH (test with litmus) and add KCN solution. Now add 
(NH^S. Compare with copper. 

(e) Potassium iodide. 

(f) Potassium chromate. 


88 


Name 


Date 


Section No. 


CADMIUM SALTS 


(a) Cd(N0 3 ) 2 + H 2 S = 

CdS + H 2 S0 4 = 

CdS + HNOs = 

(b) Cd(N0 3 ) 2 + H 2 S0 4 = 

(c) Cd(N0 3 ) 2 + NH 4 OH = 

(d) Cd(N0 3 ) 2 + NH 4 OH + KCN = 

(e) Cd(N0 3 ) 2 + KI = 

(f) Cd(N0 3 ) 2 + K 2 Cr0 4 = 

(a) CdS0 4 + H 2 S = 

(b) CdS0 4 + H 2 S0 4 = 

(c) CdS0 4 + NH 4 OH = 

(d) CdS0 4 + NH 4 OH + KCN = 

(e) CdS0 4 + KI = 

(f) CdS0 4 + K 2 Cr0 4 = 
Cd(NH 3 ) 4 S0 4 + H 2 S = 
Cd(NH 3 ) 4 (N0 3 ) 2 + H 2 S = 
Cd(CN) 2 + (NH 4 ) 2 S = 


89 




































































































































































EXERCISE 8.— ARSENIC 


Note. — Arsenic forms compounds in two different valences, corresponding to the two oxides, 
As 2 0 3 and As 2 0 5 . Both oxides are acidic and form salts analogous to those formed by the oxides 
of phosphorus; e.g., Na 3 As0 3 sodium arsenite, and Na 3 As0 4 sodium arsenate. As 2 0 3 , however, 
reacts slightly with hydrochloric acid, giving AsC 1 3 in solution, the arsenites and also arsenates, 
when in hydrochloric acid solution, will react with hydrogen sulfide as if the chloride of arsenic 
were present in the solution. Therefore, arsenic, in whatever compound it is present in the solution, 
will be precipitated with Group II. 

1. Try the action of each of the following reagents on a solution of sodium arsenite slightly 
acidified with hydrochloric acid: 

(a) Hydrogen sulfide. 

(b) Sodium hydroxide. 

(c) Ammonium hydroxide. 

(d) Dilute sulfuric acid. 

(e) Mercuric chloride. 


92 


Name 


Date 


Section No. 


ARSENIOUS SALTS 


(a) Na 3 As 03 -l - H 2 S — 

(b) Na 3 As0 3 + NaOH = 

(c) Na 3 As0 3 + NH4OH = 

(d) Na 3 As0 3 + H 2 S0 4 = 

(e) Na 3 As0 3 + (NH 4 ) 2 S = 

As 2 S 3 T" (NH 4 ) 2 Sx = 

H 3 As 0 3 + NH 4 OH + NH 4 C1 + MgS0 4 = 

















o 






















































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r 






































EXERCISE 9.— ANTIMONY 


1. Try the action of each of the following reagents on a solution of antimony chloride or 
nitrate: 

(a) Hydrogen sulfide. Test the solubility of the precipitate in dilute HC1; also in warm 
con. HC1. 

(b) Sodium hydroxide. 

(c) Ammonium hydroxide. 

(d) Ammonium sulfide. 

(e) Water. Is the precipitate soluble in HC1? 


96 


Name 


Date 


Section No 


ANTIMONY SALTS 


(a) 

SbCl 3 + h 2 s = 


Sb 2 S 3 "b dil. HC1 = 


Sb 2 S 3 + warm con. HC1 = 

(b) 

SbCl 3 + NaOH = 

(c) 

SbCl 3 + NH 4 OH = 


SbCl 3 + (NH 4 ) 2 S = 

(e) 

SbCl 3 + H 2 0 = 


Sb(OH) 3 + HC1 = 

(a) 

Sb(N0 3 ) 3 + H 2 S = 

(b) 

Sb(N0 3 ) 3 + NaOH = 

(c) 

Sb(N0 3 ) 3 + NH 4 OH> 

(d) 

Sb(N0 3 ) 3 + (NH 4 ) 2 S = 

(e) 

Sb(N0 3 ) 3 + H 2 0 = 


Sb 2 S 3 + (NH 4 ) 2 Sx = 


97 








p 









































































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EXERCISE 10. —TIN 


Note. — Tin forms two series of compounds, stannous and stannic salts. Both sulfides 
are insoluble in dilute hydrochloric acid, and hence, if tin is present in either valence it is pre¬ 
cipitated with Group II. 

1. Try the action of each of the following reagents on a solution of stannous chloride, and also 
on a solution of stannic chloride: 

(a) Hydrogen sulfide. Test the solubility of the precipitate in each case in dilute HQ; 
also in warm con. HQ. 

(b) Sodium hydroxide. Try excess of sodium hydroxide. 

(c) Ammonium hydroxide. 

(d) Ammonium sulfide. 

(e) Mercuric chloride. 


100 


Name 


Date 


Section No, 


TIN SALTS 

(a) SnCL + H 2 S = 

SnS + dil. HC1 = 

SnS + warm con. HC1 = 

(b) SnCl 2 + NaOIi = 

(c) SnCl 2 + NH 4 OH = 

(d) SnCl 2 + (NH 4 ) 2 S = 

(e) SnCl 2 + HgCl 2 = 

(a) SnCL + H 2 S = 

(b) SnCL + NaOH = 

(c) SnCL + NH4OH = 

(d) SnCL + (NH 4 ) 2 S = 

(e) SnCL + HgCl 2 = 

SnS + (NH 4 ) 2 Sx = 

SnS 2 -f" (NH 4 ) 2 Sx = 


101 




































































































































EXERCISE 11.— IRON 


1. Try the action of each of the following reagents on a solution of ferrous sulfate, and also 
on a solution of ferric chloride: 

(a) Ammonium hydroxide. Note carefully the color of the precipitate when first formed 
when the reagent is added to ferrous sulfate solution. 

(b) Repeat (a), first adding considerable ammonium chloride. 

(c) Sodium hydroxide. See if the precipitate will dissolve in an excess of the reagent. Try 
boiling it. 

(d) Ammonium sulfide. Compare the precipitate. Add HC1 to each. Write equations. 

2. (a) To a little ferric chloride solution add hydrochloric acid and then pass H 2 S gas through 

the solution for a few minutes. What is the precipitate? What is the condition of the 
iron in the solution? Test for ferric iron by adding ammonium thiocyanate. This 
experiment illustrates what takes place in the analysis of a mixture containing ferric 
compounds in the precipitation of Group II. 

(b) Potassium ferrocyanide is used to test for iron in the ferric condition. Try its action 
on a solution of ferric chloride which has been slightly acidified with hydrochloric acid. 
For comparison try its action on a solution of ferrous sulfate in the same way. 

(c) Potassium ferricyanide is used to test for iron in the ferrous condition. Try its action 
on acidified ferrous sulfate solution; also on ferric chloride for comparison. 

(d) Acidify a little ferric chloride with acetic acid and add sodium acetate solution. A 
reddish solution results. Boil. The iron is completely precipitated as a basic acetate. 


104 


Name 


Date 


Section No. 


IRON SALTS 


(a) ' FeS0 4 + NH4OH = 

(b) FeS0 4 + NH4CI + NH4OH = 

(c) FeSOi + NaOH = 

Fe(OH) 2 + NaOH = 

(d) FeS0 4 + (NH 4 ) 2 S = 

FeS + HC1 = 

FeS0 4 + NH4CNS = 

FeS0 4 + K 4 Fe(CN) 6 = 
FeS0 4 -I - K 2 Fe(CN)6 = 

(a) FeCls + NH4OH = 

(b) FeCls + NH4CI + NH4OH = 

(c) FeCls + NaOH = 

Fe(OH) 3 + NaOH = 

(d) FeCls + (NH 4 ) 2 S = 

Fe 2 S 3 + HC1 = 

FeCls + NH4CNS = 

FeCL + K4Fe(CN)s = 
FeCls + K 3 Fe(CN)a = 


105 

































































































EXERCISE 12.— ALUMINUM 


1. Try the action of each of the following reagents on a solution of aluminum sulfate; 

(a) Ammonium hydroxide. 

(b) Repeat (a), first adding considerable ammonium chloride. 

(c) Sodium hydroxide. Add the reagent slowly, noting carefully the effect of an excess. 
Boil the solution. 

(d) Ammonium sulfide. What is the precipitate? What gas is evolved? Explain. 


108 


Name 


Date 


Section No. 


ALUMINUM SALTS 


(a) A1 2 (S0 4 ) 3 + NH4OH = 

(b) A1 2 (S0 4 ) 3 + NH 4 C1 + NH 4 OH = 

(c) A1 2 (S0 4 ) 3 + NaOH = 

Al(OH) 3 + NaOH = 

(d) A1 2 (S0 4 ) 3 + (NH 4 ) 2 S = 

A1C1 3 + nh 4 oh = 

A1C1, + NaOH = 

A1C1 3 + (NH 4 ) 2 S = 

A1C1 3 + Na 2 C0 3 = 

Al(OH) 3 + HC1 = 


109 


























































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EXERCISE 13.— CHROMIUM 

1. Try the action of each of the following reagents on a solution of chromic sulfate, or chromic 
chloride: 

(a) Ammonium hydroxide. 

(b) Repeat (a), first adding a considerable amount of ammonium chloride. 

(c) Sodium hydroxide. When an excess has been added to the cold solution, boil the 
solution. 

# 

(d) Ammonium sulfate. 


112 


Name 


Date 


Section No, 


CHROMIUM SALTS 

(a) Crj(S 04)3 + NH 4 OH = 

(b) Cr 2 (S 0 4 )3 + NH 4 CI + NH 4 OH = 

(c) CrCl 3 + NaOH = 

Cr(ONa ) 3 + H 2 0 + heat = 

(d) Cr 2 (S04)3 + (NH 4 ) 2 S = 

(a) CrCla + NH 4 OH = 

Cr(NH 3 ) 4 Cl 3 + H 2 0 + heat = 

(b) CrCb + NILCl + NH 4 OH = 

(c) CrCL + NaOH = 

(d) CrCl 3 + (NH 4 ) 2 S = 

Cr(OH ) 3 + KNO 3 + Na 2 C0 3 (fuse sohds) = 
K 2 Cr 2 07 d - HC1 = 

Na 2 Cr0 4 + (CH 3 COO) 2 Pb = 

Cr(OH ) 3 + HC1 = 


113 



































































































* 














EXERCISE 14. —COBALT 


1. Try the action of each of the following reagents on a solution of cobaltous nitrate: 

(a) Hydrochloric acid and hydrogen sulfide. 

(b) Ammonium hydroxide. 

(c) Repeat (b), first adding considerable ammonium chloride. 

(d) Ammonium chloride, ammonium hydroxide, and ammonium sulfide. Is ppt. soluble 
in dilute HC1? 

(e) Hydrochloric acid and nitroso-/3-naphthoI. 

(f) Borax bead test on cobalt sulfide precipitate. 


116 


Name 


Date 


Section No. 


COBALT SALTS 


(a) Co(NOs)j -f- HC1 + H 2 S — 

(b) Co(N0 3 ) 2 + NH4OH = 

(c) Co(N0 8 ) 2 + NH 4 CI + NH 4 OH = 

(d) Co(N0 3 ) 2 + NH4CI + NH4OH + (NH 4 ) 2 S = 

(e) Co(NO s ) 2 + HC1 + CioH 6 (NO)(OH) = 

CoS + HC1 = 

CoS + HNO3 = 

CoS + 3HC1 + HN0 3 (Aqua Regia) = 

(f) CoS + Na 2 B 4 0 7 = 

CoCl 2 + HC1 + H 2 S = 

CoCl 2 + NH4OH = 

CoCl 2 + NH4CI + NH4OH = 

CoCl 2 + NH4CI + NH4OH + (NH 4 ) 2 S = 


117 












































9 










































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ft 























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EXERCISE 15.— NICKEL 

1. Try the action of each of the following reagents on a solution of nickel nitrate: 

(a) Hydrochloric acid and hydrogen sulfide. 

(b) Ammonium hydroxide. 

(c) Repeat (b), first adding considerable ammonium chloride solution. 

(d) Am monium chloride, ammonium hydroxide, and ammonium sulfide. Is ppt. soluble 
in dilute HC1? 

(e) Two or three drops of dimethylglyoxime in slightly alkaline solution. 

(f) Borax bead test on nickel sulfide precipitate 


120 


Name 


Date. . 


Seetion No. 


NICKEL SALTS 


(a) Ni(N0 3 ) 2 + HC1 + H 2 S = 

(b) Ni(N0 3 ) 2 + NH4OH = 

(c) Ni(N0 3 >2 + NH4CI + NH4OH = 

Ni(OH) 2 + NH4CI + NH4OH = 

(d) Ni(N 0 3 ) 2 + NH4CI + NH4OH + (NH 4 ) 2 S = 

(e) Ni(N0 3 ) 2 + CH 3 (C(NOH)) 2 CH 3 = 
Ni(NH 3 ) 4 (N0 3 ) 2 + H 2 S = 

Ni(NO,) 2 + (NH 4 ) 2 S = 

NiS + HC1 = 

NiS + HN0 3 = 

NiS + 3HC1 + HN0 3 (Aqua Regia) = 

(f) NiS + Na 2 B 4 0 7 (fused) = 

NiCl 2 + HC1 + H 2 S = 

NiCl 2 + NH4CI + NH 4 OH + (NH 4 ) 2 S = 


121 







, 











EXERCISE 16. — MANGANESE 

1. Try the action of each of the following reagents on a solution of manganous sulfate: 

(a) Hydrochloric acid and hydrogen sulfide. 

(b) Ammonium hydroxide. 

(c) Repeat (b), first adding considerable ammonium chloride solution. 

(d) Ammonium chloride, ammonium hydroxide, and ammonium sulfide. Is ppt. soluble 
in dilute HC1? 

(e) Sodium hydroxide. Note the color of the precipitate when first formed and the subse¬ 
quent changes upon standing. 

(f) Dry Pb0 2 and dilute HN0 3 . Boil. Red lead or sodium bismuthate may be used. 

(g) Borax bead test on sulfide ppt. 


124 


Name 


Date 


Section No, 


MANGANESE SALTS 


(a) MnS0 4 + HC1 + H 2 S = 

(b) M 11 SO 4 + NH4OH = 

(c) MnS0 4 + NH4CI + NH4OH = 

(d) MnS0 4 + NH4CI + NH4OH + (NH 4 ) 2 S = 

(e) MnS0 4 + NaOH = 

(f) M11SO4 + Pb0 2 + HN0 3 = 

M 11 SO 4 + Pba0 4 + HNOs = 

(g) MnS + Na 2 B 4 07 = 

MnS + KNO 3 + Na 2 C0 3 (fuse solids) = 
MnS0 4 + (NH 4 ) 2 S = 

MnS + HC1 = 


125 









EXERCISE 17. —ZINC 


1. Try the action of each of the following reagents on a solution of zinc sulfate or zinc chloride: 

(a) Hydrochloric acid and hydrogen sulfide. 

(b) Ammonium hydroxide. Add the reagent very slowly, a drop at a time until an excess 
has been added. Compare with copper. 

(c) Repeat (b), first adding considerable ammonium chloride solution. 

(d) Ammonium chloride, ammonium hydroxide, and ammonium sulfide. Is ppt. soluble 
in dilute HC1? 

(e) Sodium hydroxide. Add the reagent slowly, drop by drop until an excess has been 
added. Compare with aluminum. 


128 


Name 


Date 


Section No. 


ZINC SALTS 


(a) ZnS0 4 + HC1 + H 2 S = 

(b) ZnS0 4 + NH 4 OH = 

(c) ZnS0 4 + NH 4 C1 + NH 4 OH = 

(d) ZnS0 4 + NH 4 C1 + NH 4 OH + (NH 4 ) 2 S = 
Zn(NH 3 ) 4 S0 4 + (NH 4 ) 2 S = 

(e) ZnS0 4 + NaOH = 

Zn(OH) 2 + NaOH = 

ZnS0 4 + HiS = 

ZnS + HC1 = 

ZnS + HN0 3 = 

Zn(NH 3 )4S0 4 + H 2 S = 

(a) ZnCl 2 + HC1 + H 2 S = 

(b) ZnCl 2 + NH 4 OH = 

(c) ZnCl 2 + NH 4 C1 + NH 4 OH = 

(d) ZnCl 2 + NH 4 C1 + NH 4 OH + (NH 4 ) 2 S = 
Zn(NH 3 ) 4 Cl 2 + (NH 4 ) 2 S = 

(e) ZnCl 2 NaOH = 

ZnCl 2 + H 2 S = 

Zn(NH 3 ) 4 Cl 2 + H 2 S = 


129 












% 

















































































* 




























I 














































































. 










EXERCISE 18. —BARIUM 

1. Try the action of each of the following reagents on a solution of barium chloride: 

(a) Hydrochloric acid, hydrogen sulfide, ammonium chloride, ammonium hydroxide, 
and ammonium sulfide. Note the effect as each reagent is added. 

(b) Ammonium chloride, ammonium hydroxide, and ammonium carbonate. 

(c) Potassium dichromate. Test the solubility of the precipitate in acetic acid. 

(d) Ammonium oxalate. Test the solubility of the precipitate in acetic acid. 

(e) Dilute sulfuric acid. Test the solubility of the precipitate in dilute HC1. 

(f) Calcium sulfate solution. 

(g) Ammonium sulfate. 

(h) Sodium carbonate. Is ppt. soluble in dilute HC1? Compare with (e). 


132 


Name 


Date 


Section No 


BARIUM SALTS 

(a) BaCl 2 + HC1 + H 2 S + NH4CI + NH4OH + (NLDaS = 

(b) BaCl 2 + NH4CI + NH4OH + (NH 4 ) 2 C 0 3 = 

(c) BaCl 2 -f- K 2 Cr 2 0 7 = 

BaCr0 4 + CH3COOH = 

(d) BaCl 2 + (COONH 4 ) 2 = 

(COO) 2 Ba + CH3COOH = 

(e) BaCl 2 -j- HaSCL = 

BaS0 4 + HC1 = 

(f) BaCl 2 + CaS0 4 = 

(g) BaCl 2 -f- (NH 4 ) 2 S 0 4 = 

(h) BaCl 2 + Na 2 C0 3 = 

BaC0 3 T - HC1 = 

BaHP0 4 + HC1 = 

(COO) 2 Ba + HN 0 3 = 

BaC0 3 + NaHC0 3 = 


133 




































































































I 


































EXERCISE 19. — STRONTIUM 

1. Try the action of each of the following reagents on a solution of strontium chloride: 

(a) Hydrochloric acid, hydrogen sulfide, and ammonium sulfide. Note the effect as each 
reagent is added. 

(b) Ammonium chloride, ammonium hydroxide, and ammonium carbonate. 

(c) Potassium dichromate. Test the solubility of the precipitate in acetic acid. (Compare 
with barium.) 

(d) Ammonium oxalate. Test the solubility of the precipitate in acetic acid. 

(e) Dilute sulfuric acid. Test the solubility of the precipitate in dilute HCL 

(f) Calcium sulfate solution. 

(g) Ammonium sulfate. 

(h) Sodium carbonate. 


136 


Name 


Date 


Section No 


STRONTIUM SALTS 

(a) SrCl 2 + HC1 + H 2 S + (NH 4 ) 2 S = 

(b) SrCl 2 + NH4CI + NH4OH + (NH 4 ) 2 C 0 3 = 

(c) SrCl 2 -(- K 2 Cr 2 0 7 = 

SrCr0 4 + CH3COOH = 

(d) SrCl 2 + (COONH 4 ) 2 = 

(COO) 2 Sr + CH3COOH = 

(e) SrCl 2 + H 2 S0 4 = 

SrS0 4 + HCI = 

(f) SrCl 2 + CaS0 4 = 

(g) SrCl 2 + (NH 4 ) 2 S0 4 = 

(h) SrCl 2 + Na 2 C0 3 = 


137 





















f 




















- 






» 














• l 








































. 



















































. 


























EXERCISE 20.— CALCIUM 

1. Try the action of each of the following reagents on a solution of calcium chloride: 

(a) Hydrochloric acid, hydrogen sulfide, and ammonium sulfide. Note the effect as 
each reagent is added. 

(b) Ammonium chloride, ammonium hydroxide, and ammonium carbonate. 

(c) Potassium dichromate. 

(d) Ammonium oxalate. Test the solubility of the precipitate in acetic acid. (Compare 
with barium and strontium.) 

(e) Dilute sulfuric acid. 

(f) Calcium sulfate solution. 

(g) Ammonium sulfate. 

(h) Sodium carbonate. 


140 


Name 


Date 


Section No 


CALCIUM SALTS 


(a) CaCl 2 + HC1 + H 2 S + (NH 4 ) 2 S = 

(b) CaCl 2 + NH4CI + NH4OH + (NH 4 ) 2 C 0 3 = 

(c) CaCl 2 + K 2 Cr 2 0 7 = 

(d) CaCl 2 + (COONH 4 ) 2 = 

(COO) 2 Ca + CH3COOH = 

(e) CaCl 2 + H 2 S0 4 = 

(f) CaCl 3 + CaS0 4 = 

(g) CaCl 2 + (NH 4 ) 2 S0 4 = 

(h) CaCl 2 + Na 2 C0 3 = 

CaCl 2 + NaOH = 

(COO) 2 Ca + HC1 = 


141 




















» 


» 














































































EXERCISE 21. — MAGNESIUM 

1. Try the action of each of the following reagents on a solution of magnesium sulfate: 

(a) All of the group reagents of the five groups, adding them in turn to the same solution. 
Be sure to add considerable ammonium chloride before making alkaline with ammo¬ 
nium hydroxide. 

(b) Ammonium chloride, ammonium hydroxide, and ammonium oxalate. (Compare with 
calcium.) 

(c) Ammonium hydroxide. Test the solubility of any precipitate which forms in ammo¬ 
nium chloride. 

(d) Sodium hydroxide. 

(e) Sodium carbonate. Test the solubility of the precipitate in ammonium chloride. 

(f) Ammonium sulfate. (Compare with Group V.) 

(g) Ammonium chloride, ammonium hydroxide, and disodium hydrogen phosphate. Let 
stand a few minutes or rub the inside of test tube with a glass rod. 

2. (a) To a solution of magnesium sulfate add potassium hydroxide. Are the results the 

same as when sodium hydroxide is used? 

(b) Add dilute sulfuric acid to a solution of magnesium sulfate. 

(c) Precipitate a little magnesium ammonium phosphate. Test its solubility in dilute 
mineral acids. Is it soluble in acetic acid? In ammonium hydroxide? In sodium 
hydroxide? 


144 


Name 


Date 


Section No 


MAGNESIUM SALTS 

(a) MgS0 4 + HC1 + H 2 S + NH 4 C1 + NH 4 OH + (NII 4 ) 2 S + (NH 4 ) 2 C0 3 = 

(b) MgS0 4 + NH 4 C1 + NH 4 OH + (COONH 4 ) 2 = 

(c) MgS0 4 + NII 4 OH = 

Mg(OH ) 2 + NH 4 C1 = 

(d) MgS0 4 + NaOH = 

(e) MgS0 4 + Na 2 C0 3 = 

MgC0 3 + NH 4 C1 = 

(f) MgS0 4 + (NH 4 ) 2 S0 4 = 

(g) MgS0 4 + NH 4 C1 + NH 4 OH + Na 2 HP0 4 = 

MgNH 4 P0 4 + HC1 = 

MgNH 4 P0 4 + HNOa = 

MgNH 4 P0 4 + II 2 S0 4 = 

MgNH 4 P0 4 + CH 3 COOH = 

MgNH 4 P0 4 + NH 4 OH = 

MgNH 4 P0 4 + NaOH = 


145 























































* 





































































































































♦ 

























EXERCISE 22.— SODIUM 


1. (a) Moisten a little solid sodium chloride with hydrochloric acid, bring a clean platinum 
wire in contact with it, and heat it in the flame of a Bunsen burner. Repeat, examining 
flame through a blue glass. 

(b) Acidify a little sodium chloride solution with acetic acid and add sodium cobaltic 
nitrite. Warm the mixture and let it stand a few minutes. 

EXERCISE 23. — POTASSIUM 

1. (a) Moisten a little solid potassium chloride with hydrochloric acid, bring a clean platinum 
wire in contact with it, and heat in the flame of the Bunsen burner. Notice the violet 
color imparted to the flame. Repeat, examining flame through a blue glass. 

(b) Acidify a little potassium chloride solution with acetic acid and add sodium cobalt 
nitrite. Warm the mixture and let it stand a few minutes. 

EXERCISE 24. — AMMONIUM 

1. (a) Make a list of all the soluble ammonium compounds you have used in the first thirty 

exercises. 

2. Try the action of each of the following reagents on a solution of ammonium chloride: 

(a) Sodium hydroxide. (Warm.) 

(b) Potassium hydroxide. 

(c) Calcium sulfate solution. 


148 


Name 


Date 


Section No. 


SODIUM SALTS 


(a) NaCl + Na 3 C0(N0 2 ) 6 = 

NaCl + Na 3 C0(N0 2 ) 6 + CH 3 COOH = 

POTASSIUM SALTS 

(b) KC1 + Na 3 C0(N0 2 ) 6 = 

KC1 + Na 3 C0(N0 2 ) 6 + CH 3 COOH = 

AMMONIUM SALTS 

(a) NH 4 C1 + NaOH = 

(b) NH4CI + KOH = 

(c) NH4CI + CaS0 4 = 

NH 4 CI + Na 3 C0(N0 2 ) 6 = 

NH 4 C1 solid heated = 

NH 4 N0 3 solid heated = 

(NH 4 ) 2 S0 4 solid heated = 

NH 4 C1 + CaO solid heated = 


149 















PART III 


QUALITATIVE SEPARATION AND ANALYSIS 
OF METALLIC AND ACID IONS 













































































PREPARATION OF THE SOLUTION FOR ANALYSIS 

If the unknown sample is a liquid, it should be diluted to about four to six times its original 
volume, with distilled water before proceeding with the scheme of analysis. Always reserve about 
half your solution for the second analysis in case your first analysis should be lost through careless 
or incorrect procedure. 

If the sample is a solid, grind to a powder and boil a small portion in distilled water. If it does 
not dissolve, add dilute HC1 and boil. If this does not dissolve all the residue, treat another small 
portion of the solid sample with concentrated HNO 3 or aqua regia. (Aqua regia is made by using 
three volumes of concentrated HC1 to one volume of concentrated HNO 3 .) If the HC1 or HN0 3 
dissolve part of the sample, this should be filtered, and the filtrate combined with other filtrates 
obtained in a similar manner. If much HC1 is used the filtrate should be evaporated almost to 
dryness and diluted. If HNO 3 is used, evaporate the solution just to dryness, dilute with 
water, add a few drops of HC1, filter, and proceed with the analysis. If HNO 3 or aqua regia 
do not dissolve a solid, — insoluble sulfate, oxides, or silicates are probably present. These com¬ 
pounds require fusion with Na 2 C 03 in a crucible. After fusion dissolve in HC1, filter, and pro¬ 
ceed as above when HC1 was the original solvent. 


NOTES AND CAUTIONS IN PREPARATION OF SOLUTION FOR ANALYSIS 

1 . If much HC1 is used, some of the sulfides of Group II might not precipitate out with H 2 S. 

2. If sample is completely soluble in cold, dilute HC1, the metals of Group I are absent. 

3. If nitric acid is present and the solution evaporated to dryness and overheated, some soluble 
nitrates are converted to insoluble oxides. In some cases a white precipitate is formed even when 
not overheated. This probably indicates antimony as Sb 2 0 3 or tin as metastannic acid. Such 
precipitates should be heated in a partially covered crucible with excess of Na 2 C 03 and fe. The 
sulfostannate and sulfantimoniate formed are soluble in water and when treated with HC1 and 
H 2 S will be precipitated as sulfides. 

4. HNO 3 must be completely removed before passing in H 2 S in Group II on account of the 
fact that the H 2 S will be oxidized by HN0 3 to water and sulfur. The precipitated sulfur will 
cover the other precipitates and interfere with the analysis. Sulfur may be identified by the odor of 
S0 2 when burned. 

5. In testing for acid radicals do not overlook the fact that you may have added that acid in 
putting the solid in solution. Metals and alloys, of course, will not show acid radicals. 


153 


TESTS FOR SOME OF THE COMMON ACID RADICALS 

Unlike the metals, the acid radicals cannot conveniently be identified in a single solution. In 
general, it may be said that an individual test in a fresh portion of the substance must be made for 
each acid radical. However, the absence of certain groups of acid radicals may be shown by simple 
preliminary tests. 

Note. — A solution of the sodium salts of the acids is best suited for making tests for the acid 
radicals. If the unknown substance has been found to contain compounds of the heavy metals 
(by the “basic” analysis), or if it is a solid substance insoluble in water, the student should consult 
a textbook for directions by which to prepare such a solution. 


THE ACID GROUPS 

Group A — H 2 S, HC1, HBr, HI, or salts of these acids. 

Acids whose silver salts are insoluble in dilute nitric acid. 

Group B — H 2 S 0 4 , H 2 SO s , H2CO3, H3PO4, or salts of these acids. 

Acids whose barium salts are insoluble in neutral solution. 

Group C — HN 0 3 , HC2H3O2, or salts of these acids. (Soluble group.) 

Students will test the acids or salts of these acids separately and in groups according to follow¬ 
ing directions: 

PRELIMINARY TESTS FOR PRESENCE OF THE ACID GROUPS 

Group A — Acidify slightly a portion of the solution with dilute nitric acid. Add silver nitrate 
in excess. If a precipitate forms, make sure that it does not dissolve when a little more dilute nitric 
acid is added. If no permanent precipitate forms, Group A is absent. 

Group B — Make a portion of the solution neutral. (If the solution is acid, add dilute ammo¬ 
nium hydroxide; if it is alkaline, add dilute nitric acid.) Add barium chloride in excess. If no 
precipitate forms, Group B is absent. 

If a precipitate forms which wholly dissolves on the addition of hydrochloric acid, sulfates 
are absent. 

The behavior of the substance during the basic analysis may show the presence or absence of 
certain acid radicals. If the unknown solution is purple and manganese was found, a permanga¬ 
nate is indicated. If the solution is yellow or orange and chromium was found, a chromate or 
dichromate is indicated. 

If the solution is neutral or alkaline and arsenic was found, an arsenite or an arsenate is 
indicated. 

Further, the presence of some metals under certain conditions may be taken as evidence of the 
absence of certain acid radicals, e.g.: If the unknown substance is in a neutral solution and is 
found to contain barium, then Group B must be absent. If the solution is slightly acid and con¬ 
tains silver, then Group A must be absent. 

If a gas is evolved when hydrochloric acid is added (Group I), it should be identified. If C0 2 
is given off it indicates a carbonate, H 2 S a sulfide, S0 2 a sulfite, etc. 

All of the acid radicals that have not been shown to be absent must be identified. 

Group C — Acid radicals must be tested individually in original solution. 


154 


SPECIFIC ACID RADICAL TESTS 

After the preliminary group test has determined in which group the radical is found, individual 
tests are run as follows: 


GROUP A 

(1) Sulfide. — To a portion of the solution add cadmium sulfate solution in slight excess. 
A yellow ppt. indicates a sulfide. It a white ppt. forms, it is not to be considered as indicating a 
sulfide. If a ppt. forms, filter and use the filtrate for the tests for chloride, bromide, and iodide, 
as directed in the next sections (2), (3), (4). 

(2) Acidify a portion of the solution (from which sulfides have been removed as directed 
above) with dilute nitric acid, and add silver nitrate. If no ppt. forms, chlorides, bromides, and 
iodides are absent. If there is a white ppt. readily soluble in dilute ammonium hydroxide, a 
chloride is indicated. A nearly white, slightly cream-colored ppt. soluble only in a large excess of 
ammonium hydroxide indicates a bromide. A yellow ppt. practically insoluble in ammonium 
hydroxide mdicates an iodide. If a ppt. is obtained, further tests should be made, as indicated in 
the following sections (3), (4). 

(3) Iodide and Bromide. — Take a portion of the solution (from which sulfides have been 
removed) in a test tube and add a few drops of carbon bisulfide or chloroform. Add one drop of 
chlorine water and shake. If the chloroform is colored red, a bromide is present and iodides are 
absent. If the chloroform is colored violet, an iodide is present. In the latter case continue adding 
chlorine water a drop at a time, shaking after each addition until the violet color has apparently 
reached a maximum. Discard the chloroform and replace it with fresh chloroform. Add more 
chlorine water a drop at a time as before. A red color in the chloroform shows the presence of a 
bromide. 

Note. — The presence of a small amount of iodine does not interfere with the test for a bro¬ 
mide, but if much iodine is present, the compound which it forms with chlorine may give a reddish 
color to the chloroform and be confused with bromine. 

(4) Chloride. — If the test described in the last section (3) shows the absence of bromide and 
iodides, the silver nitrate test (2) shows conclusively the presence of chlorides. If bromides or 
iodides are present, chlorides must be tested for in another portion of the solution fiom which 
sulfides have been removed (1). Make it slightly alkaline with sodium carbonate and evaporate 
to dryness. Mix the residue with a little solid potassium dichromate. Put the mixture in a hard 
glass test tube fitted with a delivery tube. Moisten with concentrated sulfuric acid. Heat gently, 
collecting the fumes in a little sodium hydroxide or ammonium hydroxide solution. If this solu¬ 
tion becomes yellow in color, a chloride is indicated. Acidify with dilute sulfuric acid and add 
a few drops of hydrogen peroxide. A deep blue color shows the presence of a chloride. 


GROUP B 

(5) Carbonate. — To a portion of the substance add dilute sulfuric acid. If there is effer¬ 
vescence, pass the gas through the lime water. A white precipitate in the lime water shows the 
presence of a carbonate. A simple test is to suspend a drop of lime water on a glass rod or loop in 
the gas. Milky coloration indicates carbonates. 

(6) Phosphate. —Acidify a portion of the solution with nitric acid and add an excess of ammo¬ 
nium molybdate solution. Warm, but do not boil. A yellow precipitate shows a phosphate. 
Arsenates and silicates interfere with the test. Remove arsenic with hydrogen sulfide as in the 


155 


analysis of Group 2. Remove silicates by evaporating to dryness with nitric acid, dissolving m a 
little water and nitric acid and filtering. Chlorides, which may partially prevent the precipitation 
of the ammonium phosphomolybdate, may be removed by evaporating once or twice with nitric 

acid. 

(7) Sulfite. — Acidify a portion of the substance with hydrochloric acid. An effervescence 
of sulfur dioxide indicates a sulfide. The sulfur dioxide may be recognized by its odor or by its 
bleaching action on a drop of potassium permanganate solution brought into the gas on a glass rod. 

(8) Sulfate. — Acidify a portion of the solution with hydrochloric acid. Add barium chloride. 
A white precipitate indicates a sulfate. 


GROUP C 

(9) Nitrate. — To a portion of the solution in a test tube add an equal volume of concentrated 
sulfuric acid. Incline the tube and pour carefully on top of the mixture some ferrous sulfate 
solution. A brown ring where the two solutions meet shows the presence of a nitrate. Bromides 
and iodides interfere with the test. If iodides or bromides or both are present, add silver sulfate 
solution in slight excess, filter, evaporate the solution somewhat if necessary, and test as usual. 

(10) Acetate. — Add to a portion of the solution a little alcohol, and about an equal volume of 
concentrated sulfuric acid. Warm gently if necessary. The odor of ethyl acetate shows the 
presence of an acetate. A second test for acetates is the formation of a red solution of ferric 
acetate when ferric chloride is added to a neutral solution. When heated, a precipitate is thrown 
down. 


156 


NOTES AND CAUTIONS IN SEPARATION OF GROUPS OF METALS 


Cautions — Group I 

This group comprises the metals whose chlorides are precipitated by hydrochloric acid in the 
presence of dilute nitric acid, viz: lead, silver, and univalent mercury. 

If bismuth or antimony is present, their oxychlorides may be precipitated when hydrochloric 
acid is first added, but these are dissolved when more acid is added. 

Lead. — Owing to the partial solubility of lead chloride in water, it may not be precipitated in 
this group if the lead is present in small amount, or if the volume of the solution is large. It will 
be detected in the analysis of Group II, if present, whether it is found in Group I or not. 

Silver. — When the quantity of mercurous chloride is large and the quantity of silver chloride 
is small, and ammonia is added to separate the mercurous chloride from the silver chloride, it is 
possible that most or all of the silver chloride may be reduced to metallic silver by the mercury in 
the black mixture. In this case no re-precipitation of silver chloride will occur when nitric acid 
is added to the filtrate. Heat the black mixture (Hood) until the mercury and its compounds are 
volatilized; the residue may then be dissolved in nitric acid and the solution tested for silver by 
the usual method. 


Cautions — Group II, A 

Group II as a whole comprises those metals whose sulfides are precipitated from hydrochloric 
acid solution by hydrogen sulfide. As these metals are rather numerous the group is divided 
into two subdivisions, according to the behavior of the sulfides toward ammonium polysulfide. 
Group II, A, consisting of bivalent mercury, lead, bismuth, copper, and cadmium, includes 
those metals whose sulfides are insoluble in ammonium polysulfide. 

In precipitating the group the solution should not be too strongly acid or some cadmium may 
be held in the solution. Any considerable amount of free nitric acid should be avoided. 


Cautions — Group II, B 

Subdivision B of Group II consists of the metals, arsenic, antimony, and tin. Their sulfides 
are soluble in ammonium polysulfide and are thus separated from subdivision A. 

Because of the difficulty of completely removing arsenic from solutions containing arsenates, 
the procedure in precipitation in Group II is varied somewhat when that element is presumably 
present. The solution is first made strongly acid and saturated while hot with hydrogen sulfide 
gas. This is the best condition for removing the arsenic, but it holds a part at least of the cad¬ 
mium and tin in solution. The solution is then diluted until the acid concentration is low enough 
to permit the precipitation of these metals, and again saturated with hydrogen sulfide gas. 

In the separation of subdivisions A and B, the quantity of ammonium polysulfide used 
should be regulated by the quantity of the precipitates. If the precipitate is very large, it is some¬ 
times advisable to treat it twice with ammonium polysulfide, combining the filtrates to be exam¬ 
ined for subdivision B. 

A preliminary test with a small portion of the precipitate (warming it with ammonium poly¬ 
sulfide) will show whether or not only one subdivision is present. If the precipitate is entirely 
soluble, subdivision A is absent. If there is an insoluble residue, filter and acidify the filtrate 
with hydrochloric acid. If only sulfur is precipitated, subdivision B is absent. If only one sub¬ 
division is present, the treatment with ammonium polysulfide may be omitted. 


157 


Cautions — Group III 

This group comprises those metals whose hydroxides are precipitated by amm'onium hydroxide 
in the presence of ammonium chloride, viz.: iron, aluminum, and chromium. 

Iron. — Iron must be in the ferric condition, as ferrous hydroxide is not completely precipitated 
in the presence of ammonium salts. The hydrogen sulfide used in the precipitation of Group II 
reduces any iron present to the ferrous condition. It is oxidized before precipitating Group III 
by using nitric acid. The hydrogen sulfide must be completely boiled off before adding the nitric 
acid. If this is not done, some sulfuric acid may be formed and barium (Group V), if present, 
will be precipitated. 

If iron is found, its condition (ferrous or ferric) in the “unknown” solution must be tested in a 
portion of the original solution. The solution to be tested with potassium ferrocyanide or with 
potassium ferricyanide should be acidified with a small amount of a mineral acid. In testing for 
ferric compounds with ammonium thiocyanate, nitric acid should not be present, since oxides of 
nitrogen form a red compound with ammonium thiocyanate. The solution should be slightly 
acidified with hydrochloric or sulfuric acid. 

Al umin um. — If silicates are present, silicic acid may interfere with the test for aluminum, 
since it resembles aluminum hydroxide. It may be readily distinguished by its insolubility in 
nitric acid, in which aluminum hydroxide is easily soluble. 

If the precipitate of hydroxide of Group III is large, the test for the three metals are con¬ 
veniently made in separate portions as directed. If, however, it is quite small, test for ferrous and 
ferric iron in a portion of the original solution; boil all of the precipitate of hydroxides with 
sodium hydroxide, filter, test the filtrate for aluminum and the residue for chromium as directed. 

Chromium. — In the chromium test the solution of the fusion must be acidified sufficiently 
with acetic acid to prevent the precipitation of lead carbonate or aluminate (if the aluminum 
has not been previously removed). If the fusion is distinctly yellow in color, it is a sufficient indi¬ 
cation of the presence of chromium. 

If phosphates are present, the analysis of this and the succeeding groups must be varied some¬ 
what. 

Analysis When Phosphates Are Present 

When phosphates are present in the substance to be analyzed, the procedure must be varied 
as follows: 

The analysis of Groups I and II is carried out in the usual way. In testing for Group III 
when the solution is made alkaline with ammonium hydroxide, phosphates of all the metals in the 
solution except the alkalies may be precipitated together with the hydroxides of iron, aluminum, 
and chromium. Boil and filter. Call the filtrate “Filtrate A.” It is to be analyzed for all the 
succeeding groups in the usual way. 

Dissolve the precipitate in hydrochloric acid. Add ferric chloride in excess. Evaporate the 
solution to a small volume to remove most of the free hydrochloric acid. Nearly neutralize with 
sodium carbonate. Dilute with about 350 cc. of water, add about 25 cc. of sodium acetate solution, 
and 2 cc. of acetic acid. The solution should have a distinctive reddish color, showing that an 
excess of ferric chloride is present. Boil the solution for two or three minutes. This will precipitate 
all of the phosphates as ferric phosphate, and the excess of iron will be precipitated as basic ace¬ 
tates or hydroxides. Aluminum and chromium, if present, precipitate with the iron. Filter. Call 
the filtrate “Filtrate B.” Test the precipitate for aluminum and chromium in the usual way. 
Test Filtrate B in the usual way for Groups IV and V, and for magnesium, but not for the alkali 
metals. 

In the analysis of “Filtrate A” and “Filtrate B,” the precipitate obtained on the addition of 
ammonium sulfides (Group IV) may be united and analyzed together; likewise the precipitates 


158 


obtained on adding ammonium carbonate (Group V) may be united and analyzed together. The 
filtrates, however, must be analyzed separately. 

If the magnesium is found in Filtrate A, it need not be tested in Filtrate B, and vice versa. 
If it is not found in Filtrate A, it should, however, be tested in Filtrate B. 

Iron must be tested in a portion of the original substance. Make the test for both ferrous and 
ferric compounds in the usual way. 


Cautions — Group IV 

This group comprises the metals not included in the preceding groups whose sulfides are 
precipitated by ammonium sulfide from an ammoniacal solution, viz.: cobalt, nickel, manganese, 
and zinc. 

The ammonium sulfide should be added slowly and only in slight excess because of its sol¬ 
vent effect on nickel sulfide. 

In separating manganese and zinc from cobalt and nickel it must be remembered that cobalt 
and nickel sulfides will dissolve appreciably in hydrochloric acid if the acid is warm, if it is too 
concentrated, or if the treatment is long continued. 

In the borax bead test for nickel, the presence of a mere trace of cobalt will impart a blue color 
to the bead. The cobalt must, therefore, be completely removed before the nickel can be detected 
by the bead test. 


Cautions — Group V 

This group comprises those metals not included in the preceding groups whose carbonates are 
precipitated by ammonium carbonates in the presence of ammonium chlorides, viz: barium, 
strontium, and calcium. 

The filtrate, after removing Group IV, should be acidified with hydrochloric acid and boiled 
without delay until free from hydrogen sulfide, as long exposure to the air may result in the 
oxidation of some sulfide to sulfuric acid or a sulfate, which, in case barium is present in the solu¬ 
tion, would result in the precipitation of barium sulfate. 


Cautions — Group VI 

Group VI, sometimes called “Soluble Group,” comprises the metals which are not precipitated 
with the first five groups, viz.: magnesium, and the alkali metals — sodium, potassium, and am¬ 
monium. 

The ammonium radical must always be tested in the “original solution,” since ammonium 
salts are added at various times during the separation of the preceding groups. 

The flame test should not be relied upon for potassium, but may be used to confirm the test 
with sodium cobaltic nitrite. 


159 


ANALYSIS OF A KNOWN OR UNKNOWN SOLUTION WHICH MAY CONTAIN 
ANY OR ALL OF THE METALS 


Group I 

Note. — See sample outline of separation which immediately follows this discussion of analy¬ 
sis of metals. 

(1) To the solution add dilute hydrochloric acid as long as a precipitate forms. Filter and 
wash the precipitate with a small amount of cold water. As the succeeding groups may be present, 
the filtrate must be tested with a few drops of hydrochloric acid to be sure of complete precipitation. 
Save this filtrate to use for Group II. 

(2) Lead. — Transfer the precipitate and paper to a beaker, add water, boil and filter, catching 
the filtrate in a clean test tube. Divide it into two portions. To the first portion add a few drops 
of dilute sulfuric acid. A white precipitate indicates the presence of lead. This test is more deli¬ 
cate if a little alcohol is added to the solution. To the second portion add a few drops of potassium 
chromate solution. A yellow precipitate, soluble in an excess of sodium hydroxide, shows the 
presence of lead. 

If desired, other tests used in Exercise 1 may be used to confirm the presence of lead. 

If lead is present, wash the precipitate remaining on the filter with hot water until the washings 
do not give a precipitate with sulfuric acid. 

(3) Silver. — Pour over the precipitate a little dilute ammonium hydroxide, collecting the 
filtrate in a clean beaker. Make the filtrate slightly acid with nitric acid. A white precipitate 
shows the presence of silver. 

(4) Mercury(ous). — If the precipitate blackens when the ammonium hydroxide is added, 
mercury is present. 


Group II 

(1) Heat the filtrate from Group I in a small Erlenmeyer flask nearly to boiling and pass into 
it under pressure hydrogen sulfide gas for about 2 to 5 minutes. Reheat the solution and again pass 
in the hydrogen sulfide gas for about 2 minutes. It should now be thoroughly saturated and the 
precipitate should settle rapidly, leaving the solution clear on top. (Complete saturation may be 
rapidly determined by placing the hand over the mouth of the flask and shaking the liquid. If a 
suction is felt on the hand, the solution is not yet saturated with the gas.) Let the precipitate 
settle and filter while hot. Dilute the filtrate with 20 to 25 cc. of water, heat, and again satuiate 
with hydrogen sulfide gas to be sure of complete precipitation. As the succeeding groups may 
be present, the filtrate should be saved for Group III. AA ash the precipitate with hot water until 
it is free of chlorides. To determine this, catch a little of the washings on a small watch glass and 
add one drop of silver nitrate solution. A white turbidity indicates chlorides. I he washing should 
be continued until this test gives a negative result. 

(2) Transfer the precipitate to a small beaker or evaporating dish, add to it 4 to 10 cc. of 
ammonium polysulfide solution and warm, but do not boil, for five to ten minutes. 4 he mixture 
should be stirred frequently during this treatment. Dilute with an equal volume of water and 
filter. If there is a residue it contains a sulfide of Group II, A. 4 he filtrate may contain Group 
II, B, arsenic, antimony, and tin in the form of their ammonium sulfo-salts. Save this filtrate 
to test for metals of Group II, B. 


160 


Group II, A 

(3) Mercury(ic). — Transfer the precipitate to a small beaker or evaporating dish and add a 
little dilute nitric acid (dilute the concentrated acid with about four volumes of water for this 
purpose.) Roil gently for a short time. This should dissolve all but the mercuric sulfide. Dilute 
and filter. Test the filtrate as directed in the next section (4). Transfer the residue to a test tube, 
add a little concentrated hydrochloric acid and 3 or 4 drops of concentrated nitric acid, and boil. 
Dilute the solution and filter. Boil this solution until all free chlorine is expelled, cool, and add 
stannous chloride solution, at first drop by drop and finally in excess. A white or grayish precipi¬ 
tate which becomes darker with excess of stannous chloride, shows that mercury is present. 

(4) Lead. — Add to the filtrate from the mercuric sulfide 3 to 4 cc. of concentrated sulfuric 
acid and evaporate in a porcelain dish until dense white fumes appear. Cool the liquid and pour 
it carefully into a small beaker one-half full of water. Stir and let stand a few minutes. A white 
precipitate is lead sulfate. Filter and test the filtrate as directed in the next section (5). Wash 
the precipitate and then pour repeatedly over it about 10 cc. of ammonium acetate solution. 
Acidify the solution thus obtained with acetic acid and add potassium chromate solution. A 
yellow precipitate soluble in excess of sodium hydroxide shows the presence of lead. 

(5) Bismuth. — Add ammonium hydroxide to the filtrate obtained above from the lead 
sulfate until the solution smells strongly of ammonia. A white precipitate indicates bismuth. 
It is BiOOH. Stir thoroughly, filter, and wash the precipitate. Test the filtrate as directed in the 
next section (6). Pour over the precipitate a little freshly prepared sodium stannite solution. If 
the precipitate blackens, bismuth is present. (Note. — Sodium stannite is prepared by adding 
sodium hydroxide solution to a solution of stannous chloride until the precipitate which forms just 
dissolves.) 

(6) Copper. — If the solution obtained on adding ammonium hydroxide in section (5) is blue, 
copper is present. Boil off the excess of ammonia and test the solution as directed in the next 
section (7). 

(7) Cadmium. — If copper is absent, acidify the solution slightly with hydrochloric acid and 
saturate it with h»ydrogen sulfide gas. A yellow precipitate shows the presence of cadmium. 

If copper is present, shown by the deep blue color, add solution of potassium cyanide, enough 
to give a colorless solution and add solution of ammonium sulfide. A yellow precipitate shows 
that cadmium is present. (Caution: Potassium cyanide is a deadly poison. Be absolutely certain 
that the solution is alkaline to litmus with ammonium hydroxide before adding the cyanide. If 
acid, a deadly gas, hydrogen cyanide, will be produced. When through with this test, run water 
in sink and wash the test tube of solution through the drain with plenty of water.) 

Another method of testing for cadmium, if copper is present, is as follows: Acidify the solution 
slightly with hydrochloric acid and completely precipitate both the copper and the cadmium (if 
present) with hydrogen sulfide gas. Filter and at once boil the precipitate with 6-N sulfuric acid. 
This dissolves the cadmium sulfide but not the copper sulfide. Dilute with an equal volume of 
water, filter, and saturate the filtrate with hydrogen sulfide gas. A yellow precipitate shows the 
presence of cadmium. 


Group II, B 

(1) To the ammonium polysulfide filtrate add dilute hydrochloric acid until it is distinctly 
acid. (Test with litmus paper.) Sulfur will separate as a milk-white precipitate, and arsenic, 
antimony, and tin, if present, will form flocculent, colored precipitates. If only sulfur is precipi¬ 
tated, Group II, B is absent. If a flocculent or colored precipitate forms, filter, discarding the 
filtrate. 


161 


(2) Arsenic. — Add just 10 cc. of concentrated hydrochloric acid to the precipitate and heat 
for a few minutes. Dilute with a very little water and filter. Wash the residue with a little water, 
adding the washings to the filtrate. Add enough water to the filtrate to make the total volume 
just 50 cc. and test it for antimony and tin as directed in the next section (3), (4), (5). Transfer 
the residue of arsenic sulfide and sulfur to a test tube and boil it with a little concentrated 
nitric acid. Boil off most of the excess of nitric acid, dilute, and filter if necessary to obtain a clear 
solution. Make the solution alkaline with ammonium hydroxide, then add ammonium chloride 
and magnesium sulfate solution. If a precipitate does not form at once, rub the inside of the test 
tube with a glass rod and let it stand for a few minutes. A white crystalline precipitate shows the 
presence of arsenic. If desired, the nitric acid solution may be tested for arsenic with silver ni¬ 
trate, making a neutral zone with ammonium hydroxide. 

(3) Antimony. — The solution obtained from (2) should have a volume of just 50 cc. and should 
contain just 10 cc. of concentrated hydrochloric acid. Saturate it hot with hydrogen sulfide gas. 
Under these conditions the antimony is precipitated and the tin is held in solution if these metals 
are present. An orange-colored precipitate shows the presence of antimony. Filter if there is a 
precipitate. Add 5 cc. of water to the solution and again saturate with hydrogen sulfide gas to 
be sure that the antimony is completely removed. Filter if necessary. Test the solution for tin. 

(4) Tin. — Dilute considerably the solution obtained from (3) and saturate it with hydrogen 
sulfide gas. A yellow precipitate shows that tin is present. To confirm this test evaporate to a 
small volume without filtering and add a piece of pure zinc. A spongy deposit is metallic tin. 
Wash it by decantation and dissolve it by warming with a little con. HC1. Dilute the solution 
slightly and add it at once to a solution of mercuric chloride. A white or gray precipitate shows 
the presence of tin. 

(5) Another method of detecting tin and antimony in the hydrochloric acid solution obtained 
from the sulfides is as follows: Dilute the solution. Put a piece of pure zinc on a piece of plati¬ 
num foil and lower it into the solution. The solution should be diluted sufficiently for a slow evolu¬ 
tion of hydrogen. A black stain on the platinum foil, which is insoluble in sodium hypochlorite, 
indicates that antimony is present; then an excess of zinc is added (about one gram), the tin is 
allowed to separate and is dissolved in concentrated hydrochloric acid and identified by the test 
with mercuric chloride, as described above. Antimony may separate out with the tin but it is 
insoluble in hydrochloric acid. 


Group III 

(1) Boil the filtrate obtained from Group II until it is free from H 2 S. This may be tested by 
the odor or by holding in the steam a bit of filter paper moistened with lead acetate solution. Add 
a few drops of nitric acid and boil again. Add excess of solid ammonium chloride. (Ammonium 
chloride is formed in the solution also when ammonium hydroxide is added.) Keep the solution 
boiling and add slowly, with constant stirring, ammonium hydroxide until the solution smells 
strongly of ammonia. (A large excess of ammonium hydroxide must be avoided.) Dissolve the 
precipitate in nitric acid, re-precipitate with ammonium hydroxide and filter. As the succeeding 
groups may be present, the filtrate should be saved for Group IV. If the filtrate smells strongly of 
ammonia, precipitation may be assumed to be complete. Wash the precipitate with hot water. 

(2) Iron. — Dissolve a portion of the precipitate in hydrochloric acid. Dilute the solution and 
divide into two portions. To the first portion add ammonium thiocyanate. A blood-red color 
shows the presence of iron. To the second portion add potassium ferrocyanide solution. A deep 
blue color shows the presence of iron. 


162 


(3) Aluminum. — Boil a second portion of the precipitate with sodium hydroxide solution. 
Filter. Acidify the filtrate with hydrochloric acid, and then make it slightly alkaline with ammo¬ 
nium hydroxide. A white precipitate soluble in nitric acid shows the presence of aluminum. 

Another simple test for any aluminum salt is to heat the solid on a piece of charcoal with a 
blowpipe and while hot touch with a drop of some cobalt solution. Heat again. A robin-egg blue 
color indicates aluminum. 

(4) Chromium. — Fuse a third portion of the precipitate in a clean porcelain dish with a few 
crystals of potassium nitrate and some dry sodium carbonate. Cool. A yellow color indicates 
chromium. If the yellow color is not easily apparent, dissolve the fusion with a little hot water, 
filter, acidify the solution with acetic acid, and add lead acetate solution. A yellow precipitate, 
soluble in an excess of sodium hydroxide, shows the presence of chromium. 


Group IV 

(1) Heat to boding the filtrate obtained after removing Group III. Add ammonium hydroxide, 
and then with frequent stirring add slowly ammonium sulfide until an excess is present. Boil 
off the excess, then add a few drops more making a slight excess. Boil, let the precipitate settle, 
filter while hot, and test the filtrate with a drop of ammonium sulfide to be sure of complete pre¬ 
cipitation. As the succeeding groups may be present, the filtrate must be saved for Group V. 
The precipitate may contain CoS, NiS, MnS, and ZnS. 

(2) Wash the precipitate with hot water. Pierce the filter paper with a stirring rod and wash 
the precipitate into a beaker with cold water. Use enough water to completely cover the precipi¬ 
tate in the beaker. Add a very little concentrated hydrochloric acid. (With the water in the 
beaker this forms dilute acid. The volume of concentrated acid added should be about 1/10 
or 1/12 of the volume of the water in the beaker.) Stir the mixture thoroughly and filter it at 
once. A black residue may be CoS, or NiS, or both of these. The filtrate may contain manganese 
and zinc. 

(N ote . — If the group precipitate is very light-colored, with no suggestion of brown or black, 
cobalt and nickel are absent. The precipitate may then be completely dissolved in a small amount 
of dilute HC1, and the solution tested for manganese and zinc.) 

(3) Cobalt and Nickel. — Wash the black residue of cobalt and nickel sulfides and dissolve 
the residue in a little aqua regia. (If the quantity of precipitate is very small, boil the paper and 
residue together with aqua regia until the black sulfides have all dissolved, dilute somewhat, and 
filter.) Evaporate the solution just to dryness, but do not ignite the residue, and then dissolve 
this residue in a small amount of water (about 10 cc.). Divide in two portions. 

(a) An excellent test for nickel is made by making alkaline with ammonium hydroxide and 
adding to this solution a little dimethylglyoxime. A red precipitate of nickel dimethylglyoxime 
will be formed if nickel is present. This test can be made in the presence of any other metal. 

(b) Add 3-4 cc. of nitroso-jS-naphthol. A brick red precipitate indicates cobalt. 

(4) Manganese. — Boil the filtrate containing the manganese and zinc thoroughly, and make 
it strongly alkaline with sodium hydroxide. If manganese is present, a precipitate of the hydroxide 
is formed. This precipitate is light-colored at first but rapidly darkens by oxidation to a brownish 
color. Filter and test the filtrate for zinc as directed in the next section (5). Transfer a little of 
the precipitate to a clean porcelain dish, add a little dry sodium carbonate and a few small crystals 
of potassium nitrate, and fuse the mixture. Cool. A green residue shows the presence of man¬ 
ganese. 

A good test for manganese on the original solution is to add 5 cc. dilute HNO 3 , warm, add a 


163 


little red lead or lead dioxide or sodium bismuthate, warm, and allow to stand. A purple-colored 
solution indicates manganese. 

(5) Zinc. — Make the filtrate obtained after removing the manganese slightly acid with acetic 
acid, and then add ammonium sulfide. A white precipitate shows the presence of zinc. 


Group V 

(1) Acidify the filtrate obtained after removing Group IV with dilute hydrochloric acid, and 
boil it down to a volume of about 30 to 35 cc. Filter if necessary to obtain a clear solution, as 
some sulfur may separate during the evaporation. Add ammonium chloride, and make the 
solution slightly alkaline with ammonium hydroxide. Have the solution boiling, add ammonium 
carbonate to complete precipitation, stir, and let it stand a few minutes. Filter and wash the 
precipitate. The filtrate must be saved for Group VI. The precipitate may contain BaC0 3 , 
SrC0 3 , CaC0 3 . 

(2) Dissolve the precipitate in a little acetic acid, add sodium acetate, boil, add excess K 2 Cr 2 0 7 . 
Filter off yellow precipitate of BaCr0.j. (This yellow precipitate is the test for barium.) To the 
filtrate add an excess of ammonium hydroxide and ammonium carbonate. V arm one minute. 
Filter and wash the precipitated carbonates of calcium and strontium. Dissolve in the least 
possible amount of HC1, evaporate just to dryness and re-dissolve in about 4 cc. of water. The 
chlorides of calcium and strontium are formed. Filter if necessary and divide into two portions. 
To the first portion add CaS 04 solution and boil. Let stand 20 minutes until a white precipitate 
of strontium sulfate forms. (This is the test for strontium.) To the second portion add a 
saturated solution of ammonium sulfate and bcil. Let stand for 30 minutes and filter. All the 
strontium is in the precipitate and enough calcium is in the filtrate for a test. To the filtrate add 
ammonium oxalate and warm. A white precipitate of calcium oxalate forms. (This is the test for 
calcium.) 

Group VI 

(1) To the filtrate obtained after removing Group V, add a few drops of ammonium sulfate 
and ammonium oxalate to remove traces of barium, strontium, and calcium. Notice carefully 
whether either reagent gives any precipitate. Filter and discard any precipitate which may form. 
Divide the solution into two unequal parts and test the smaller portion for magnesium (2), and 
the larger portion for the alkali metals (3). 

(2) Magnesium. — Add to the smaller portion of the solution ammonium hydroxide and 
sodium phosphate. If a precipitate does not form at once, rub the inside of the test tube with a 
stirring rod and let it stand for a few minutes. A white crystalline precipitate shows the presence 
of magnesium. 

(3) The Alkali Metals. — Evaporate the larger portion containing the alkali metals to dryness 
in an evaporating dish and ignite the residue to expel ammonium salts. 1 hese are given off in the 
form of white fumes, and the heating must be continued until no more fumes are evolved. 

Moisten a portion of the residue with a few drops of hydrochloric acid, and introduce some of 
it into the Bunsen flame on a clean platinum wire. A violet flame shows the presence of potassium 
when viewed through a blue glass. 

A bright yellow flame shows the presence of sodium. 

Very small quantities of sodium compounds are readily detected by the flame test. The test 
is not so delicate for potassium. The presence of sodium will completely mask the potassium flame. 

Viewed through the spectroscope, potassium gives two red lines and a violet line. The latter 
is in the extreme violet end of the spectrum and is difficult to see. One of the red lines is also 
quite faint. Sodium gives a strong yellow line. If the sodium line is faint or appears only in 

1 G 4 


flashes, it is probably caused by dust in the air coming into the flame, and not by an appreciable 
amount of sodium in the substance under examination. 

Dissolve the remainder of the residue in a very little water, make the solution slightly acid 
with acetic acid, and add sodium cobaltic nitrite solution. Warm the mixture and let it stand for a 
time. A yellow precipitate shows the presence of potassium. 

(Note. — It is essential that ammonium salts be completely removed before making this test 
for potassium, since they also give a precipitate with this reagent.) 

(4) Amm onium. — The ammonium radical must be tested in the original substance. To a 
small portion of the original solution in a small beaker add sodium hydroxide. Cover the beaker 
with a watch glass, on the under side of which a piece of moist red litmus paper has been placed. 
Warm the solution almost to the boiling point. If the litmus paper becomes blue it shows the 
presence of the ammonium radical. The evolution of ammonia may also be confirmed by its 
characteristic odor. 


165 


SAMPLE OUTLINE OF SEPARATION OF GENERAL UNKNOWNS 


In running a general unknown always reserve about half of the original solution to use in case 
determination is lost. 


(Make your own outline for practice.) 


Group I 


Soluble Salts of Pb, Ag, Hg, such as AgN0 3 , Pb(N0 3 ) 2 , PIgN0 3 . 
To unknown solution add Dil. HC1 and Filter. 


[Residue.. 

AgCl, PbCl 2 HgCl (wash, transfer to beaker) and add H 2 O, 
boil and filter. 


.. .Filtrate 

Groups II, III, IV, V, VI, 

(Save) 


T 


|Residue. 

AgCl, HgCl 

Wash with hot H 2 0 
add 

NH 4 OH on filter 

. . . Filtrate Divide in 3 portions 

(1) PbCl 2 + H 2 S0 4 = PbS0 4 (White) 

(2) PbCl 2 + K 2 Cr0 4 = PbCr0 4 (Yellow) 

(Soluble in NaOH) 

(3) PbCl 2 + KI = Pbl 2 (Yellow) 

|Residue. 

.Filtrate 

HgNH 2 Cl + Hg 

Ag(NH 3 ) 2 Cl 

(black on filter paper) 

add (until acid) 


HN0 3 = AgCl (White) 


Note. — If there is no white Ppt. when HC1 is added, there are no Group I metals present. 


166 











Group II 

Note. — Before running in IBS, drop blue litmus into filtrate from Group I. It will turn red. 
Then just make alkaline with NH 4 OH. Add HC1 to acidify and proceed as outlined. 

Filtrate from Group I may be PbCl 2 , CuCl 2 , CdCl 2 , BiCl 3 , HgCl 2 , AsC 1 3 , SbCl 3 , SnCl 2 , add 
dilute HC1; pass in H 2 S as long as Ppt. forms. Filter. 


(Residue...Filtrate 


PbS, CuS, CdS, Bi 2 S 3 , HgS, As 2 S 3 , Sb 2 S 3 , SnS, SnS 2 , transfer 
to beaker and add (NH 4 ) 2 Sx; warm and filter 


Warm, test with H 2 S to insure 
precipitation is complete. Groups 
III, IV, V, VI. (Save) 


|Residue.Filtrate (B) 


(Wash with water to remove (NH 4 ) 2 Sx). PbS, 
CuS, CdS, Bi 2 S 3 , HgS. Transfer paper and residue 
to beaker 
add 

1:1 HN0 3 , warm and filter 


See next page. (Save) 


(Residue.Filtrate 


HgS, transfer paper and residue to beaker and 
add Aqua Regia (3HC1 + HN0 3 ), evaporate nearly 
to dryness; add a little warm H 2 0 and filter. 

This filtrate is HgCl 2 (Sol.). Add SnCl 2 which 
forms SnCl 4 + Hg (Black) 


Pb(N0 3 ) 2 
Cu(N 0 3 ) 2 
Cd(N0 3 ) 2 
Bi(N0 3 ) 3 
add H 2 S0 4 ; 


evaporate until white fumes 


dilute and filter 


T 


(Residue.Filtrate 


PbS0 4 
add 

H 2 0, warm and pour into large 
volume of cold H 2 0. Milky Ppt. 
shows Pb present as PbS0 4 sol¬ 
uble in ammonium acetate. 


CuS0 4 , CdS0 4 , Bi 2 (S0 4 ) 3 
add 

NH 4 OH and filter 


Residue.Filtrate (Two portions) 


Bi(OH) 3 on filter paper add 
HC1, allow to drop through into 
a large beaker of H 2 0 if White 
Ppt. (Bi) as BiOCl. 

Sodium stannite test may be 
substituted. 


A. (1) If Cu is present the 
solution Cu(NH 3 ) 4 S0 4 will be 
blue. Another test, treat blue 
solution as follows: 

(2) Acidify with acetic acid, 
then addK 4 Fe(CN)6 = 
Cu 2 Fe(CN) 6 Brick Red. Cu. 

B. (1) Cd(NH 3 ) 4 S0 4 if blue 
color add excess of NH 4 OH (test 
with litmus) add KCN solution 
(very poisonous) till blue color 
disappears. Add (NH 4 ) 2 S = 
CdS yellow. 

(2) If colorless use only 
(NIB) 2 S. 


167 


















Filtrate B 


Add Dil. HC1 till acid. (Milky Ppt. = S.) If Ppt. is colored, As, Sb, Sn may be present. 
F ilter and discard filtrate. 

Residue: add 10 cc. con. HC1, warm and filter. Wash with little water on filter. 


[Residue...Filtrate 


Transfer to test tube and boil with 
little con. HN0 3 . Boil off excess HN0 3 , 
dilute and filter. Make alkaline with 
NH 4 OH, add NH 4 CI and MgS0 4 . 
White crystalline Ppt. = As. 


Make up to 50 cc. Volume. Saturate while hot with 
H 2 S until complete precipitation and filter. 


Residue.Filtrate 


Orange Ppt. = Sb. 


Dilute with H 2 0, warm and 
pass in H 2 S. Yellow Ppt. = Sn. 


Group III 


Take filtrate from Group II and boil off the H 2 S (test with filter paper moistened in Pb(N0 3 ) 2 ) 
— then add a few drops of con. HN0 3 and an excess of solid NH 4 C1; warm, add slight excess 
NH 4 OH; filter and wash the residue with warm II 2 0. 


[Residue. 

Fe(OH ) 3 

Al(OH ) 3 Divide in three portions. 
Cr(OH ) 3 


.Filtrate 

Groups IV, V & VI (Save). 


(1) First portion: add Dil. HC1 until Ppt. dissolves = FeCl 3 and pour into two test tubes. 

a. Add K 4 Fe(CN ) 6 deep blue ppt. = Fe. 

b. Add NH 4 CNS, blood-red color = Fe. 

(2) Second portion: add NaOH = Na 3 A10 3 ; boil, filter t.o clean beaker. To filtrate (Na 3 A10 3 ) 
add HC1 till turns litmus red, add NH 4 OH till alkaline = Al(OH) 3 . V hite Ppt. = Al. 

Another test for Al. — heat ppt. on charcoal with blowpipe, touch with drop of cobalt solution, 
reheat. Blue color = Al. 

( 3 ) Third portion: place in evaporating dish. Add solid KN0 3 + solid Na 2 C0 3 and fuse — 
Yellow res. = Cr. If not sure let dish cool, add a little H 2 0, warm and filter; acidify with acetic 
acid, then add Pb(N0 3 ) 2 — Orange Yellow = Cr. 

Note. — When getting the first residue — if white there can only be Al present; but if dark, 
all may be present. 

If iron is present, test original unknown to determine whether ferrous or ferric. 


168 











Group IV 


Make part of filtrate from Group III alkaline with NH4OH and add (NH^S slowly, warm 
and filter. (The reason for saving a portion of the filtrate is because it will be needed in testing 
for Mn.) If no Ppt. no Group IV. 


|Residue . Filtrate (Save) 

NiS Groups V and VI 

CoS If ppt. is black, there is Ni or Co present. 

MnS If ppt. is light color, there is no Ni or Co present. 

ZnS 


Put paper and ppt. into beaker and add cold dilute HC1 and stir with glass rod and filter. 


|Rcsidti6 . 

.Filtrate 


NiS 

CoS 

(a) Test small portion with borax 
bead using platinum loop. 

(b) Balance of residue in evap. dish, 
add Aqua Regia, evaporate to dryness 

MnCl 2 

ZnCl 2 

add 

NaOH and Filter 


and dilute with H 2 0, put into test tube. 
NiCl 2 

Residue . 

.Filtrate ] 

(1) Mn(OH ) 2 put in evaporating 

Na 2 Zn0 2 acid 

CoCl 2 

dish, add solid KN0 3 and Na 2 CC >3 and 

ify with FLC 

Separate in two portions. 

fuse — Green color = Mn. 

and add 

(a) 1st Portion: make alkaline with 

(2) Test some of residue with borax 

(NH 4 ) 2 S 

NH 4 OH, add few drops of Dimethyl- 

bead. 

ZnS 

glyoxime. Red color = Ni. 

(3) Take some of the original filtrate 

White Ppt. = 

(b) 2nd Portion: add dil. HC1, add 10 
cc. H 2 0 and warm, add 3 cc. nitroso-/3- 
naphthol. Brick Red Ppt. = Co. 

of Group III and add about 5 cc. dil. 
HNO 3 , heat to boiling, hold over sink 
and add some solid PbsCb, Pb0 2 , or so¬ 
dium bismuthate. Let settle; purple 
color indicates Mn. The Mn. is in the 
form of HMnCL. 

Zn. 


169 













Group V 

Acidify filtrate from Group IV with HC1 and boil until about 30 to 35 cc. If not clear, filter. 
Then add NH 4 OH till alkaline. Boil, add (NH 4 ) 2 C0 3 , stir, let settle, filter, and wash. 


|Residue. 

BaCOa 
Sr CO 3 
CaC0 3 

Dissolve Ppt. in a little acetic acid, add sodium acetate, boil, 
add excess K 2 Cr 2 0 7 . Filter. 


. Filtrate 

Group VI (Save) 



| 

|Residue. 

.Filtrate 


Barium 
as 

BaCr0 4 Yellow 
or 

Ba Flame Test = Green. 


Add excess of NH 4 OH and (NH 4 ) 2 C 03 . Warm, filter, 
wash Ppt. on filter. 

Dissolve Ppt.with the least amount of HC1. Evaporate 
just to dryness, dissolve in about 5 cc. H 2 0, filter if not 
clear. Divide into two portions. 



First Portion. 

Add CaS0 4 and boil. If no Ppt. appears im¬ 
mediately, let stand for ten minutes. A fine 
white Ppt. proves the presence of Strontium; 
or 

Sr Flame Test = Red. 


.Second Portion 

Add (NH 4 ) 2 S0 4 and boil, let stand for 30 min¬ 
utes and filter. All the strontium is in the Ppt. 

To the Filtrate add (NH 4 ) 2 C 2 0 4 and warm. A 
white Ppt. of Calcium oxalate forms; 
or 

Ca Flame Test = Brick Red for a second, then 


Yellow. 


Note. — Flame tests should be run on a portion of the filtrate from Group IV. 


170 












Group VI 


To the filtrate from Group V, add a few drops of (NH^jSO* and (NID 2 C 2 O 4 to remove traces 
of Barium, Strontium, and Calcium. Note carefully whether either reagent gave a Ppt. Filter 
and discard any Ppt. which may have formed. Divide the solution into two (2) portions. 


First Portion. 

Add NH4OH and NasHPCb. If a Ppt. does 
not form, rub the inside of the test tube with a 
glass rod and let stand for a few minutes. A 
white ppt. (crystalline) shows presence of Mag¬ 
nesium- 

Ammonium — Take a portion of the Original 
Unknown and place in a small beaker and add 
NaOH. Cover the beaker with a watch glass, on 
the imder side of which has been placed a piece of 
moist red litmus. Warm the solution almost to 
boiling. If litmus becomes blue, there is an am¬ 
monium radical present. Do not allow solution 
to spatter upon litmus. 


.Second Portion 

Evaporate to dryness and continue to evapo¬ 
rate till no fumes are given off. This drives off 
the ammonium salts. Moisten a portion of the 
residue with a few drops of HC1, and introduce 
some of it into the Bunsen flame on a clean 
platinum wire. 

A violet flame shows the presence of Potassium 
when viewed through a blue glass. 

A Bright Yellow flame shows the presence of 
Sodium. 

Dissolve the remainder of the residue in a little 
H 2 0, make the solution slightly acid with acetic 
acid and add Na 3 Co(N0 2 )6, warm the mixture 
and let it stand for a time. Yellow Ppt. shows 
the presence of Potassium. 

Note. — If the solution is acid when the water 
is added, it should first be made neutral by the 
addition of sodium carbonate and then should 
be acidified with acetic acid. 


ACID RADICALS 
N ote. — Acid tests must be run on original unknown. 


171 




SUGGESTED REVIEW ON ACID RADICALS AND GROUP I 

1. Give equations involved and color of ppt. for 3 tests for Pb, 3 tests for Ag, and 3 tests for 
Hg. 

2. Give color of: PbCl 2 , PbCr 04 , PbS 04 , PbS, Pbl 2 . 

3. Give colors of: AgCl, Ag 2 Cr0 4 , Ag 2 S0 4 , Ag 2 S, Agl. 

4. Give colors of HgCl, Hg 2 Cr0 4 , HgS. 

5. A ppt. consists of PbCl 2 , AgCl. How separate PbCl 2 from the other? 

6. A ppt. consists of PbCr0 4 , Ag 2 Cr0 4 , Hg 2 Cr0 4 . How separate PbCr0 4 from the others? 

7. A ppt. consists of Ag 2 Cr0 4 and Hg 2 Cr0 4 . How separate Ag 2 Cr0 4 from the others? 

8. A ppt. consists of Pbl 2 and Hgl. How prove Pb is present? How prove Hg is present? 

9. Outline the separation of metals used, if Pb(N0 3 ) 2 and HgN0 3 are known to be present 
and Ag is absent. 

10. Outline the separation of metals used, if Pb(N0 3 ) 2 and AgN0 3 are known to be present 
and Hg is absent. 

11. Outline the separation of metals used, if nitrates of Pb, Ag, and Hg are suspected to be 
present. 

12. Give the simple test for chloride ion, if a solution of NaCl is used. 

13. Ditto for bromide ion if solution of NaBr is used. 

14. Ditto for iodide ion if solution of Nal is used. 

15. Ditto for sulfide ion if solution of H 2 S is used. 

16. Ditto for simple ions of Group B and C, if soluble Na salt of each acid is used separately. 

17. Original first unknown + BaCl 2 gives white ppt. insoluble in HC1, yet sulfate ion is not 
present. Explain. How must the test be run? 

18. Original first unknown + FeS0 4 -f- H 2 S0 4 gives white ppt. What is the ppt. and how 
should the nitrate test be run? 

19. Original unknown + FeCl 3 in testing for acetate gives a ppt. What is the ppt. and how 
should the test be run? 

20. The HC1 filtrate of the first group unknown is treated with FeCl 3 but fails to show red 
color, although acetate is known to be present. Why? How remedy? 

21. The HC1 filtrate of first group unknown is treated with ethyl alcohol and concentrated 
H 2 S0 4 and odor of CH 3 COOH is noted but no odor of ethyl acetate. Explain. How remedy? 

22. The HC1 filtrate of first group unknown is treated with excess of ethyl alcohol and con¬ 
centrated H 2 S0 4 is poured carefully down the tube. Acetate is known to be present, yet no odor 
of ethyl acetate is obtained. Explain. How remedy? 

23. W r hy should care be exercised in running the nitrate and acetate tests? 

24. An unknown contains both bromides and iodides. How prove each ion is present? 

25. An unknown contains both chlorides and iodides. How prove each ion is present? 

26. A solution of silver nitrate is treated with HC1 and the ppt. formed is dissolved in NH 4 OH. 
On addition of HN0 3 a white ppt. drops out. Give equations. 

27. A white ppt. formed by addition of HC1 to mercurous nitrate is treated with NH 4 OH. 
Give equations involved. What is color of ppt.? 

28. A solution containing nitrates of Pb, Ag, and Hg is treated with dilute HC1. The white 
ppt. is filtered off. To the filtrate we add HC1. A white ppt. forms. Explain. How proceed? 

29. The chloride ppt. of Group I metals is washed on filter with cold water. Why? Will the 
washings give a test for lead? 

30. The chloride ppt. of Group I metals is washed on filter with several test tubes of hot 
water. How test to prove all PbCl 2 is removed? 

31. After several washings with hot water, the Group I chloride ppt. is treated with NH 4 OH 
on filter. A milky solution comes through the filter and persists in the NH 4 OH solution. Explain. 
How proceed? 


172 


32. On addition of HN0 3 to the milky NH 4 OH solution in (31) the solution clears. What was 
the ppt.? 

33. On addition of HN0 3 to the milky NH 4 OH solution in (31) the solution fails to clear. 
What was the ppt.? 

34. A sample containing a large amount of HgN0 3 and a trace of AgN0 3 was treated according 
to outline of separation of Group I metals. On treatment of the chlorides with NH4OH no silver 
test was shown. Explain. How proceed? 

35. A sample contains a large amount of HgN0 3 and AgN0 3 but only a small quantity of 
Pb(N0 3 ) 2 . The Group I analysis fails to show Pb present. How proceed? 

36. A known sample containing all Group I metals was diluted with large volume of distilled 
water. No test for lead was obtained. How would you proceed to precipitate out the lead in 
Group I? 

37. Why do we test for lead in both Group I and Group II? 

38. Why do we test for mercury in both Group I and Group II? 

39. An unknown was found to contain lead. The original was a clear aqueous solution. 
Which of the following acid radicals would you expect to find present: sulfate, carbonate, nitrate, 
sulfide, phosphate, chloride, acetate, bromide? 

40. Same as (39) except Ag was found present. 

41. Same as (39) except Hg was found present. 

REVIEW GROUP II 

1 . Name the metals of Group II. What is the group reagent? 

2. Which of these metals are found in A and which in B division of this group? 

3. What is the purpose of subdividing this group? 

4. Why are Pb and Hg in this group as well as Group I? 

5. Name the metals whose sulfides are insoluble in dilute HC1. 

6 . Name the sulfides soluble in ammonium polysulfide. 

7. Give the color of each sulfide in the group. 

8 . Which sulfide of the group is insoluble in 50% IIN0 3 ? What advantage can be taken of 

this? 

9. Which metals of this group have salts which give a milky ppt. when diluted? 

10. Name the yellow sulfides of the group. 

11. Name the black sulfides of the group. 

12. Which is the only reddish orange sulfide in our qualitative separation? 

13. Which acid ions form insoluble salts with any or all the metals of the group? 

14. A clear unknown is treated with dilute HC1 in beginning Group II analysis. A white ppt. 
is thrown down. What is indicated? How proceed? 

15. A Group II unknown is found to be strongly acid. How proceed? 

16. Which metals of this group will not be precipitated as sulfides if solution is strongly acid 
when H 2 S is passed in? 

17. On passing in H 2 S a heavy white ppt. is formed. What is indicated? How proceed? 

18. If solution is acid, what reagent is used to neutralize? Why? 

19. If solution is alkaline, what reagent is used to neutralize? Why ? 

20. A strongly acid unknown is treated with NH4OH. Even before neutral a white ppt. is 
produced. What metals are indicated? How proceed? 

21. A clear unknown is treated with NH4OH. A deep blue color is produced. \\ hat metal is 
indicated? Will this test work in the presence of all other metals of the group? 

22 . What metals of this group will be thrown down as precipitates on addition of H 2 S? 

23. How can you tell when a hot solution in an Erlenmeyer flask is saturated with H 2 S? 

24. After passing in H 2 S, the sulfides are filtered. The filtrate is treated with more H 2 S. A 
ppt. is obtained. What is indicated? How proceed? 

173 


25. On passing in H 2 S a reddish orange or orange yellow ppt. is obtained. What metals are 
known to be absent? How proceed? 

26. After treatment of Group II sulfides with (NH 4 ) 2 Sx the filtrate is acidified with HC1. 
A white ppt. is obtained. What is the ppt. and what metals are absent? 

27. After treatment of a yellow sulfide ppt. with (NH 4 ) 2 Sx, on filtering no residue remains. 
What metals are known to be absent and what metals may be present? 

28. After treatment of a yellow sulfide ppt. with (NH 4 ) 2 Sx, on filtering a residue is left and 
treatment of the filtrate with HC1 gives a white ppt. What metals are known to be absent and 
what present? 

29. The combined sulfides of Division A are treated with 50% HN0 3 . No residue remains. 
What metal is absent? 

30. The HgS residue is boiled with aqua regia. To what form is the Hg converted? What is 
the rubbery insoluble residue? How prove? 

31. The sulfide ppt. is black, yet tests fail to show any metal present. What may be the 
trouble and how treat the solution to repeat the separation? 

32. A yellow sulfide ppt. is obtained, no Division B is present, and yet Cd fails to show. What 
may be the trouble? 

33. The mixture of copper and cadmium ammonium sulfates is warmed with pulverized 
iron until blue color disappears. The filtrate is treated with ammonium sulfide and a spongy ppt, 
obtained, but it is black. What is the difficulty? How remedy to repeat the test? 

34. The ppt. obtained from filtrate B is light yellow. What metals are absent? 

35. The ppt. obtained from filtrate B is reddish orange. What metals may be present? 

36. Explain the separation of Sb and Sn. 

37. What sulfide of filtrate B is insoluble in concentrated HC1? Which are soluble? 

38. What will be the effect if HN0 3 is present in the second unknown? How remedy? 

39. Outline the separation of Group II, assuming that all metals are present. 

REVIEW GROUP III 

1. Name the metals of Group III and the group reagent. 

2. In an unknown containing Groups I, II, and III, why are Group III metals not precipi¬ 
tated with H 2 S? 

3. If an unknown containing Groups I, II, and III were made alkaline with NH 4 OH, what 
precipitates would be thrown down? 

4. The third unknown is acidified with HC1. A white ppt. is thrown down. What is the ppt. 
and how proceed? 

5. The third unknown is treated with dilute HC1. On addition of H 2 S a ppt. is obtained. 
What is the ppt. and how proceed? 

6. In a general unknown we use the filtrate from the preceding groups. Why is it necessary 
to boil off all H 2 S from Group II filtrate before proceeding with Group III? 

7. Why is the third group unknown treated with HN0 3 ? 

8. Give color of each hydroxide of Group III metals. 

9. The NH 4 OH ppt. of Group III metals is white. What metals are absent and what present? 

10. The NH 4 OH ppt. of Group III metals is dark red. What metal is known to be present? 
May the other metals be present? 

11. The NH 4 OH ppt. of Group III metals is dark green. What metal is known to be present? 
May the other metals be present? 

12. Are all hydroxides of this group soluble in dilute HC1? 

13. Do the presence of A1 and Cr interfere with the potassium ferrocyanide or the ammonium 
thiocyanate tests for iron? 

14. The precipitated hydroxides of Fe, Al, and Cr are boiled with NaOH. Which metal is 
removed? Give equation. Do the other metals interfere? 

174 


15. What is the purpose of KN0 3 in the Cr fusion? What Cr salt is formed? 

16. This test for Cr is the reverse of a test used previously. What metal was tested by the same 
reaction? 

17. What is the purpose of the excess of NII 4 CI before precipitating the metals of this group 
with NH 4 OH? 

18. A white solid unknown is heated on charcoal and while hot is moistened with a cobalt 
solution. On further heating a blue color is obtained. What metal is indicated? 

19. In running the fusion test for Cr, the melt is dissolved in water. On addition of a Pb salt 
a white ppt. is obtained. What is the ppt.? 

20. What is the purpose of acidifying with CH 3 COOH in the Cr fusion test? 

21 . Why avoid excess of reagents in dissolving the Cr fusion? 

22 . What is the test for ferrous iron in an unknown? 

23. What is the test for ferric iron in an unknown? 

24. Does the test for iron using ammonium thiocyanate distinguish between ferrous and 
ferric iron? 

25. Why must the ammonium thiocyanate test be run in absence of HNO s ? 

26. How must our method of separation be varied if phosphates are present? 

27. What white gelatinous ppt. similar to A1 (OH) 3 may be precipitated on addition of NH 4 OH? 
How distinguish between these two ppts.? 

28. Outline the separation of the metals of Group III. 

29. An unknown contains Fe, Al, Cr in aqueous solution. What acid ions cannot be present? 
What may be present? 

30. In running the Group III unknown, why is it necessary to save the first filtrate?- 

31. The Group III unknown in solution is colorless. What metals are known to be absent? 

32. The Group III unknown in solution is dark greenish in color. What metal may be ex¬ 
pected? 

33 . The Group III unknown in solution is reddish brown in color. What metal may be 
expected? 


REVIEW GROUP IV 

1. Name the metals of Group IV and the group reagent. 

2. On addition of dilute HC1 to a solution to be tested from Group IV metals, a white ppt. 
is obtained. What is the ppt.? How proceed? 

3. Dilute HC1 and H 2 S are added to a solution to be tested for Group IV metals. A yellow 
ppt. is obtained. What is the ppt.? How proceed? 

4. On addition of NH 4 OH to a solution to be tested for Group IV metals, a ppt. is ob¬ 
tained. What is the ppt.? How proceed? 

5. Give color of each sulfide in this group. 

6 . In running a general unknown, H 2 S is passed into an acidified solution. No ppt. is ob¬ 
tained. NH4OH is added and a black ppt. is thrown down which is insoluble in ammonium poly¬ 
sulfide or 50% HN0 3 , but dissolved in aqua regia. No test for Hg is obtained. What is the ppt.? 

7. H 2 S is used in precipitation of metals of Group II and Group IV. Why are Group IV 
metals not precipitated along with Group II when H 2 S is first used? 

8 . Which sulfides of this group are soluble in dilute HC1? Which are insoluble? 

9. What would be the effect of addition of hot or concentrated HC1 to Group IV sulfides? 

10. The sulfide ppt. of this group is white. What metal is expected? Which metals are 
known to be absent? 

11. The sulfide ppt. of this group is muddy-colored. What metals may be present? 
Which absent? 

12. The sulfide ppt. of this group is black. What metals are expected? What may be present? 

13. The original aqueous solution is reddish in color. What metal is suspected? 

175 


14. The original aqueous solution is green in color. What metal is suspected? 

15. The original aqueous solution is colorless. What metals of Group IV may be present and 
what are absent? 

16. Why are the sulfides of Mn and Zn dissolved in IIC1 rather than HN0 3 ? 

17. To what form are Ni and Co sulfides converted in aqua regia? 

18. Group IV solution is green, a black sulfide is obtained, but on treating the aqua regia 
solution with dimethylglyoxime no test for Ni is obtained. Explain. 

19. An unknown gives a black ppt. with ammonium sulfide, but the sulfide is soluble in dilute 
HC1. What is the metal? 

20. Name the sulfides examined thus far which are soluble in dilute HC1. 

21. What metals examined thus far give ppts. with NH 4 OH soluble in dilute HC1? 

22. What metals examined thus far give ppts. with NaOH soluble in excess of reagent? 

23. What metals examined thus far give ppts. soluble in excess of NaOH? 

24. A borax bead test made on the black sulfide ppt. shows Co. May Ni or Mn be present? 

25. Give color of the borax bead using a Co salt; a Ni salt; a Mn salt. 

26. Describe the fusion test for Mn. 

27. Describe the HMn 04 test for Mn. 

28. What other compounds may be substituted for the PbjCh in the Mn test? 

29. MnS and ZnS are on the filter. How separate and test for each? 

30. CoS and NiS are on the filter. How‘separate and test for each? 

31. NiS and ZnS are on the filter. How separate and test for each? 

32. Outline complete separation of the metals of Group IV. 

33. The (NH 4 ) 2 S filtrate is treated with more (NH 4 ) 2 S. A ppt. forms. What is indicated? 
How proceed? 

34. In the final test for Zn the ppt. is black although Zn is known to be present. What may 
be the trouble? How proceed? 

35. A muddy-colored ppt. indicates Mn, yet the complete test fails to show the metal. How 
recover to run through the test again? 

36. An aqueous solution contains salts of the four metals of this group. What acid ions may 
be present? Which absent? 

37. Why is the aqua regia solution acidified before testing for Co with nitroso-jS-naphthol? 

38. A large excess of (NH 4 ) 2 S is added to Group IV unknown. The solution was originally 
green but no test for Ni was obtained. Explain. 

Note. — Keep in mind that it is extremely important that ALL previous metals must be re¬ 
moved before testing a filtrate for other metals. You should be able to put an unknown back in 
its original form to repeat a test any time you fail to get a test for a metal you strongly suspect 
to be present. 


REVIEW GROUP V 

1. Name the metals and the group reagent and precipitant of Group V. 

2. Why are these metals not precipitated in Group II or Group IV as sulfides? 

3. A clear, colorless aqueous unknown of salts of all these metals might contain what acid 
ions? Which ions would be absent? 

4. Give colors and comparative aqueous solubilities of the hydroxides of these metals? 

5. If the hydroxides are only slightly soluble, why does not NH 4 OH in Group III precipitate 
some of Group V metals in a general unknown? 

6 . Give colors and comparative aqueous solubilities of the sulfates of these metals. 

7. In our scheme of separation what use is made of difference in solubility of these sulfates? 

8 . Give colors of the carbonates of these metals. Are they soluble in II 2 0? In dilute HC1? 

9. In our general scheme of analysis why do we use HC1 for acidifying? 

10. Why is NH4OH used to make solutions alkaline? 


176 


11. Why is H 2 S used in Group II as the sulfide and (NH 4 ) 2 S in Group IV? 

12. Why are (NH 4 ) 2 S0 4 and (NH 4 ) 2 C0 3 and (NH 4 ) 2 C 2 0 4 used as reagents in Group V instead 
of the acids or the sodium salts? 

13. The (NH 4 ) 2 S filtrate from Group IV in a general unknown is acidified with HC1. A white 
ppt. forms. What is the ppt.? How proceed? 

14. Why is HC1 added at beginning of Group V separation? 

15. Why are the carbonates dissolved in acetic acid rather than HC1? What is the purpose 
of the sodium acetate? 

16. Which metals of this group give chromates soluble in acetic acid? Which insoluble? 

17. Why is ammonium oxalate used instead of ammonium sulfate as our final test for Ca? 

18. Describe method of running flame tests for metals of this group, and give color imparted 
to the flame by salts of each metal. 

19. What would be the result in filtrate from Group IV, containing (NH 4 ) 2 S, if allowed to stand 
for some time before running Group V metals? 

20. What metals included in the preceding groups would be precipitated with this group if not 
previously removed? 

UNKNOWN GROUP VI 

1. Name the metals of this group. 

2. Why is this called the soluble group? Will any reagent precipitate all the metals of the 
group? 

3. Why is the ammonium ion included with the metals? 

4. What acid ions may be present if only the metals of this group are present in an unknown? 

5. Why is ammonium ion tested on the original unknown? 

6. In running the ammonium test you are advised to warm the solution but not boil violently. 
Why? 

7. Before running the Na and K test the solution is evaporated to dryness and heated till 
no more white fumes are driven off. Why? 

8. What effect would the presence of NH 4 salts have on the K test? 

9. Describe the method of running flame tests for K and Na. 

10. Can a Na flame be detected in presence of K; of Ca; of Sr; of Ba? 

11. An unknown contains salts of Na, K, Ca, Sr. Describe the flame test to identify and prove 
each present. 

12. Can a K flame be detected in presence of Na, Ca, Sr, and Ba? 

13. What is the purpose of acetic acid in the cobaltic nitrite test for K? 

14. A yellow ppt. insoluble in acetic acid is obtained when testing an unknown solution, yet K 
is absent. What is the ppt.? 

15. In the Mg test what is the purpose of rubbing the inside of the test tube with a glass rod? 
What is the ppt.? 


177 


EQUATIONS 

(Precipitates are in black type) 

MERCUROUS SALTS 

HgNOa + HC1 = HNOs + HgCl 
2HgN0 3 + K 2 Cr0 4 = 2KN0 3 + Hg 2 Cr0 4 
Hg 2 Cr0 4 + NaOH = no action 
2HgN0 3 + H 2 S0 4 = 2HN0 3 + Hg 2 S0 4 
2HgN0 3 + H 2 S = 2HN0 3 + HgS + Hg 
HgN0 3 + KI = KN0 3 + Hgl 
2HgN0 3 + 2NaOH = 2NaN0 3 + H 2 0 + Hg 2 0 
Hg 2 0 =HgO + Hg 

6HgN0 3 + 6NH 4 OH = 3Hg + (NH 2 HgN0 3 ) 2 Hg0 + 4NH 4 N0 3 + 5H 2 0 
2HgN0 3 + (NH 4 ) 2 S = Hg + HgS + 2NH 4 N0 3 ■ 

6Hg + 8HN0 3 (dil.) = 6HgN0 3 + 2NO + 4H 2 0 
Hg + 4HN0 3 (conc.) = Hg(N0 3 ) 2 + 2N0 2 + 2H 2 0 
2HgCl + 2NH 4 OH = NH 4 C1 + 2H 2 0 + Hg + HgNH 2 Cl 
2HgCl + SnCl 2 = SnCl 4 + 2Hg 
HgCl + hot H 2 0 = no action 


LEAD SALTS 

(CH 3 COO) 2 Pb + 2HC1 = 2CH 3 COOH + PbCl 2 

(CH 3 COO) 2 Pb + Iv 2 Cr0 4 = 2CH 3 COOK + PbCr0 4 

PbCr0 4 + 4NaOH = Pb(ONa) 2 + Na^CrO, + 2H 2 0 

PbCr0 4 + CH 3 COOH = no action 

(CH 2 COO) 2 Pb + H 2 S0 4 = 2CH 3 COOH + PbS0 4 

(CH 3 COO) 2 Pb + H 2 S = 2CH 3 COOH + PbS 

PbS + HC1 = no action 

(CH 3 COO) 2 Pb+ 2KI = 2CH 3 COOK + Pbl 2 

(CH 3 COO) 2 Pb + 2NaOH = 2CH 3 COONa + Pb(OH) 2 

Pb(OH) 2 + 2NaOH = Pb(ONa) 2 + 2H 2 0 

(CH 3 COO) 2 Pb + NH 4 OH = no action 

(CH 3 COO) 2 Pb + (NH 4 ) 2 S = 2CH 3 COONH 4 + PbS 

Pb(N0 3 ) 2 + 2HC1 = 2HN0 3 + PbCl 2 

Pb(N0 3 ) 2 + K 2 Cr0 4 = 2KN0 3 + PbCr0 4 

Pb(N0 3 ) 2 + H 2 S0 4 = 2HN0 3 + PbS0 4 

Pb(N0 3 ) 2 + H 2 S = 2HN0 3 + PbS 

Pb(N0 3 ) 2 + 2KI = 2KN0 3 + Pbl 2 

Pb(N0 3 ) 2 + 2NaOH = 2NaN0 3 + Pb(OH) 2 

4Pb(N0 3 ) 2 + 4NH 4 OH = 2Pb0.Pb(N0 3 ) 2 + 4NH 4 N0 3 + 2H 2 0 

Pb(N0 3 ) 2 + (NH 4 ) 2 S = 2NH 4 N0 3 + PbS 

PbO + 2HN0 3 = H 2 0 + Pb(N0 3 ) 2 


178 


PbCl 2 + hot H 2 0 = PbCl 2 solution 

6 PbS + I 6 HNO 3 (dU.) = 6Pb(N0 3 ) 2 + 3S 2 + 4NO + 8H 2 0 
PbS + 8 HNO 3 (con.) = PbS0 4 + 8N0 2 + 4H 2 0 

SILVER SALTS 

AgN0 3 + HC1 = HN0 3 + AgCl 

AgCl + HNO 3 = no action 

2AgN0 3 + K 2 Cr0 4 = 2KN0 3 + Ag 2 CrO 

Ag 2 Cr0 4 + 4NH 4 OH = (Ag(NH 3 ) 2 ) 2 Cr0 4 + 4H 2 0 

2AgN0 3 + H 2 S0 4 = 2HN0 3 + Ag 2 S0 4 

2AgN0 3 + H 2 S = 2HN0 3 + Ag 2 S 

Ag 2 S + 2HNO s = H 2 S + 2AgN0 3 

AgN0 3 + KI = KNO 3 + Agl 

Agl + NH 4 OH = no action 

2AgN0 3 + 2NaOH = 2NaN0 3 + Ag 2 0 + H 2 0 

2AgN0 3 + (NH 4 ) 2 S = 2 NILNO 3 + Ag 2 S 

6 Ag -}- 8HN03(dil.) = 6 AgNOs 4 - 2NO -(- 4H 2 0 

2Ag + 4HN0 3 (con.) = 2AgN0 3 + 2H 2 0 + 2N0 2 

Ag(NH 3 ) 2 Cl + 2HN0 3 = 2NH 4 N0 3 + AgCl 

AgCl + hot H 2 0 = no action 

MERCURIC SALTS 

HgCl 2 + H 2 S = 2HC1 + HgS 

HgS + HC1 = no action 

HgCl 2 + H 2 S0 4 = no action 

HgCl 2 + 2NaOH = 2NaCl + H 2 0 + HgO 

HgCl 2 + 2NH 4 OH = NH 4 C1 + 2H 2 0 + HgNH 2 CI 

HgCl 2 + 2KI = 2KC1 + Hgl 2 

Hgl 2 + 2IvI = HgI 2 .2KI 

HgCl 2 + (NH 4 ) 2 S = 2NH 4 C1 + HgS 

HgCl 2 + K 2 Cr0 4 = 2KC1 -f- Hg 2 Cr0 4 

3HgS + 2(3HC1 + HNO 3 ) = 3HgCl 2 + 3S + 2NO + 4H 2 0 

2 HgCl 2 + SnCl 2 = SnCl 4 + 2HgCl 

2 HgCl + SnCl 2 = SnCl 4 + 2Hg 

Hg(N0 3 )2 + H 2 S = 2HN0 3 + HgS 

9 HgS + 8 HNO 3 = 3(Hg(N0 3 ) 2 + 2HgS) + 2NO + 4H 2 0 + 3S 
3(Hg(N0 3 ) 2 .2HgS) + 4(3HC1 + HNO 3 ) = 3Hg(N0 3 ) 2 + 6 HgCl 2 + 4NO + 6 S + 8H 2 0 
Hg(N0 3 ) 2 + 2KOH = 2KN0 3 + H 2 0 + HgO 

BISMUTH SALTS 

2Bi(N0 3 ) 3 + 3H 2 S = 6 HNO 3 + Bi 2 S 3 
B^Ss + HC1 = no action 

179 


2Bi(N0 3 ) 3 + H 2 SO 4 + 2H 2 0 = 6 HNO 3 + (Bi0) 2 S0 4 
Bi(N0 3 ) 3 + 3NaOH = 3NaN0 3 + Bi(OH ) 3 
Bi(N0 3 ) 3 + 3NH 4 OH = 3 NH 4 NO 3 + Bi(OH ) 3 
Bi(OH ) 3 + 3HC1 = BiCl 3 + 3H 2 0 
Bi(N0 3 ) 3 + 3KI = 3KN0 3 + Bil 3 
2Bi(N0 3 ) 3 + (NH 4 ) 2 S = 6NH4N0 3 + Bi 2 S 3 

2Bi(N0 3 ) 3 + 2K 2 Cr0 4 + H 2 0 = 4IvN0 3 + 2HN0 3 + (Bi0) 2 Cr 2 0 7 

Bi(N0 3 ) 3 + H 2 0 = 2HN0 3 + Bi0N0 3 

2BiCl 3 + 3H 2 S = 6HC1 + Bi 2 S 3 

2 BiCl 3 + H 2 S0 4 + 2H 2 0 = 6HC1 + (Bi0) 2 S0 4 

BiCl 3 + 3NaOH = 3NaCl + Bi(OH) 3 

2Bi(OH ) 3 boiled = 3H 2 0 .+ Bi 2 0 3 

Bi(OH ) 3 + 3HC1 = 3H 2 0 + BiCl 3 

BiCl 3 + 3NH 4 OH = 3NH 4 C1 + Bi(OH) 3 

BiCl 3 + 3KI = 3KC1 + Bil 3 

2BiCl 3 + 3(NH 4 ) 2 S = 6 NH 4 CI + Bi 2 S 3 

2 BiCl 3 + 2K 2 Cr0 4 + H 2 0 = 4KC1 + 2HC1 + (Bi0) 2 Cr 2 0 7 

BiCl 3 + H 2 0 = 2HC1 + BiOCl 

COPPER SALTS 

C 11 SO 4 + H 2 S = H 2 S0 4 + CuS 

CuS + H 2 S0 4 (dil.) boil = CuS0 4 + H 2 S 

CuS0 4 + H 2 S0 4 = no action 

CuS0 4 + 2 NaOH = Na 2 S0 4 + Cu(OH) 2 

Cu(OH ) 2 + heat= CuO + H 2 0 

CuS0 4 + 4NH 4 OH = Cu(NH 3 ) 4 S0 4 + 4H 2 0 

2CuS0 4 + 2Na 2 C0 3 + H 2 0 = 2Na 2 S0 4 + C0 2 + CuC0 3 Cu(0H) 2 

CuC0 3 + Cu(OH ) 2 + heat = H 2 0 + C0 2 + 2CuO 

CuS0 4 + Fe = FeS0 4 + Cu 

2CuS0 4 + 4KI = 2K 2 S0 4 + I 2 + 2CuI 

CuS0 4 + (NH 4 ) 2 S= (NH 4 ) 2 S0 4 + CuS 

CuS0 4 + K 2 Cr0 4 = K 2 S0 4 + CuCr0 4 

Cu(N0 3 ) 2 + H 2 S = 2HN0 3 + CuS 

Cu(N0 3 ) 2 + NaOH = 2NaN0 3 + Cu(OH) 2 

Cu(N0 3 ) 2 + 4NH 4 OH = 4H 2 0 + Cu(NH 3 ) 4 (N0 3 ) 2 

Cu(N0 3 ) 2 + Fe = Fe(N0 3 ) 2 + Cu 

Cu(NH 3 ) 4 S0 4 + 2H 2 S = CuS + (NH 4 ) 2 S0 4 + (NH 4 ) 2 S 

Cu(NH 3 ) 4 (N0 3 ) 2 + 2H 2 S = CuS + 2NH 4 N0 3 (NH 4 ) 2 S 

6 CuS + 16HN0 3 = 6 Cu(N0 3 ) 2 + 4NO + 3S 2 + 8H 2 0 

2CuS0 4 + K 4 Fe(CN ) 6 = Cu 2 Fe(CN) 6 + 2K 2 S0 4 

CuCr0 4 + 4NH 4 OH = Cu(NH 3 ) 4 Cr0 4 + 4H 2 0 


180 


CADMIUM SALTS 


Cd(N0 3 ) 2 + H 2 S = CdS + 2HN0 3 

CdS + H 2 S0 4 (dil.) (boil) = CdS0 4 + H 2 S 

Cd(N0 3 ) 2 + H 2 S0 4 = CdS0 4 + 2HN0 3 

Cd(N0 3 ) 2 + 2NaOH = Cd(OH ) 2 + 2NaN0 3 

Cd(N0 3 ) 2 + 4NH 4 OH = Cd(NH 3 ) 4 (N0 3 ) 2 + 4H 2 0 

Cd(N0 3 ) 2 + Na 2 C0 3 = CdC0 3 + 2NaN0 3 

Cd(N0 3 ) 2 + KI = no action 

Cd(N0 3 ) 2 + Fe = no action 

Cd(N0 3 ) 2 (NH 4 ) 2 S = CdS + 2NH 4 N0 3 

Cd(N0 3 ) 2 + K 2 Cr0 4 (boil) = CdCr0 4 + 2KN0 3 

CdS0 4 + H 2 S = CdS + H 2 S0 4 

CdS0 4 + H 2 S0 4 = no action 

CdS0 4 + 2NaOH = Cd(OH ) 2 + Na*S0 4 

CdS0 4 + 4NH 4 OH = Cd(NH 3 ) 4 S0 4 + 4H 2 0 

CdS0 4 + Na 2 C0 3 = CdC0 3 + Na 2 S0 4 

CdS0 4 + Fe = no action 

CdS0 4 + KI = no action 

CdS0 4 + K 2 Cr0 4 = CdCr0 4 + K 2 S0 4 

Cd(NH 3 ) 4 S0 4 + 2H 2 S = CdS + (NH 4 ) 2 S0 4 + (NH 4 ) 2 S 

Cd(NH 3 ) 4 (N0 3 ) 2 + 2H 2 S =CdS + 2NH 4 N0 3 + (NH 4 ) 2 S 

2CdS0 4 + K 4 Fe(CN ) 6 = Cd 2 Fe(CN) c + 2K 2 S0 4 

ARSENIOUS SALTS 

2Na 3 As0 3 3H 2 S = As 2 S 3 4- 6 NaOET 
Na 3 As0 3 + NaOH = no action unless hot 
Na 3 As0 3 + NH 4 OH = no action 
Na 3 As0 3 + Na^COs = no action 

2Na 3 As0 3 + 3H 2 S0 4 = 3Na 2 S0 4 + 2H 3 As0 3 or 3Na 2 S0 4 + 3H 2 0 + As 2 0 3 
2Na 3 As0 3 + 3(NH 4 ) 2 S = As 2 S 3 + 6 NH 4 OH if acid. As 2 S 3 dissolves if basic. 
As 2 S 3 + (NH 4 ) 2 Sx = 2(NH 4 ) 3 AsS 4 + 2S 


ANTIMONY SALTS 

. 2SbCl 3 + 3H 2 S = Sb 2 S 3 + 6HC1 red ppt. if not too acid 
Sb 2 S 3 + dil. HC1 = no action 
Sb 2 S 3 + warm con. 6HC1 = 2SbCl 3 + 3H 2 S 
2SbCl 3 + 6 NaOH = 6 NaCl + 3H 2 0 + Sb 2 0 3 ppt. 
2SbCl 3 + 3 Na 2 C 0 3 = 6 NaCl + 3C0 2 + Sb 2 0 3 ppt. 
SbCl 3 + H 2 S0 4 = no action 

2SbCl 3 + 6 NH 4 OH = 6NH 4 C1 + 3H 2 0 + Sb 2 0 3 ppt. 
2SbCl 3 + 3(NH 4 ) 2 S = Sb 2 S 3 + 6NH 4 C1 if acid 

181 


SbCl 3 + H 2 0 « SbOCl + 2HC1 

Sb(OH ) 3 + 3HC1 = SbCl 3 + 3H 2 0 

2Sb(N0 3 ) 3 + 3H 2 S = Sb,S 3 + 6HN0 3 floe, red ppt. 

2Sb(N0 3 ) 3 + 6 NaOH = Sb 2 0 3 + 6NaN0 3 + 3H 2 0 

2Sb(N0 3 ) 3 + 3Na 2 C0 3 = Sb 2 0 3 + 3C0 2 + 6NaN0 3 

Sb(N0 3 ) 3 + H 2 S0 4 = no action 

2Sb(N0 3 ) 3 + 6 NH 4 OH = Sb 2 0 3 + 3H 2 0+ 6NH 4 N0 3 

Sb(N0 3 ) 3 + (NH 4 ) 2 S = no action 

Sb(N0 3 ) 3 + 3H 2 0 = SbON0 3 + 6HN0 3 

Sb 2 S 3 + 2HN0 3 = Sb 2 0 3 + H 2 0 + 2NO + 3S 

STANNOUS SALTS 

' SnCl 2 + H 2 S = SnS + 2HC1 

SnS + dil. HC1 = in sol. (SnS is a ppt. only in dilute acid sol.) 

SnS + warm con. 2HC1 = SnCl 2 + H 2 S 

SnCl 2 + 2 NaOH = Sn(OH) 2 + 2NaCl (Sn(OH ) 2 + 2NaOH = Na Sn0 2 + 2H 2 0 sol.) 

SnCl 2 + Na 2 C0 3 + H 2 0 = C0 2 + 2NaCl + Sn(OH) 2 

SnCl 2 + H 2 S0 4 = SnS0 4 + 2HC1 

SnCl 2 + (NH 4 ) 2 S = SnS + 2NH 4 C1 

SnCl 2 + HgCl 2 = SnCl 4 + Hg 


STANNIC SALTS 

SnCl 4 + 2 H 2 S = SnS 2 + 4HC1 
SnSl 4 + 4NaOH = Sn(OH ) 4 + 4NaCl 

SnCl 4 + 4NH 4 OH = Sn(OH ) 4 + 4NH 4 C1 ppt. dissolves in excess ammonia 
SnCl 4 d - 2Na 2 C0 3 -j- 2H 2 0 = 2C0 2 4NaCl -(- Sn(OH ) 4 or 2C0 2 4NaCl H 3 Sn0 2 fl - 
H 2 0 (the Sn(OH ) 4 dissolves in excess of carbonate) 

SnCl 4 + 2H 2 S0 4 = Sn(S0 4 ) 2 + 4HC1 

SnCl 4 + 2(NH 4 ) 2 S = SnS 2 + 4NH 4 C1 

SnCl 4 + HgCl 2 = no action 

3SnS + 4HN0 3 + H 2 0 = 3H 2 Sn0 3 + 4NO + 3S 

3SnS 2 + 4HN0 3 + H 2 0 = 3H 2 Sn0 3 + 4NO + 6 S 

SnS + (NH 4 ) 2 S 2 = (NH 4 ) 2 SnS 3 

SnS 2 + (NH 4 ) 2 S = (NH 4 ) 2 SnS 3 

IRON SALTS 

FeS0 4 + 2NH 4 OH = Fe(OH ) 2 + (NH 4 ) 2 S0 4 

FeS0 4 + 2NH 4 C1 + NH 4 OH = Fe(OH ) 2 + (NH 4 ) 2 S0 4 

FeS0 4 2NaOH = Fe(OH ) 2 4 - Na 2 S0 4 

Fe(OH ) 2 4- NaOH = no reaction 

FeS0 4 4- (NH 4 ) 2 S = FeS 4- (NH 4 ) 2 S0 4 

FeS 4- 2HC1 = H 2 S 4- FeCl 2 


182 


FeSC >4 “l - Na 2 C 03 = Na 2 S 04 -f- FeC0 3 

4FeC0 3 + 0 2 + 6H 2 0 = 4Fe(OH ) 3 + 4C0 2 (2 steps) 

3FeS0 4 + 2 Na 2 HP 0 4 = Fe 3 (P0 4 ) 2 + 2NaHS0 4 + NajSO* 

2FeS + 3HN0 3 = 2Fe(N0 3 ) 2 + 2S + 4H 2 0 + 2NO 
FeS0 4 + 2 NH 4 CNS = Fe(CNS ) 2 + (NH 4 ) 2 S0 4 
2FeS0 4 + K 4 Fe(CN ) 6 = 2K 2 S0 4 + Fe 2 (Fe(CN) 6 ) 

3FeS0 4 + 2 K 3 Fe(CN ) 6 = 3K 2 S0 4 + Fe 3 (Fe(CN ) 6 ) 2 

FeCls + 3 NH 4 OH = 3 NH 4 CI + Fe(OH ) 3 

FeCl 3 + NH 4 CI + 3NH 4 OH = 4NH 4 C1 + Fe(OH ) 3 

FeCl 3 + 3NaOH = Fe(OH ) 3 + 3NaCl 

Fe(OH ) 3 + NaOH = no reaction 

4FeCl 3 + 6 (NH 4 ) 2 S = 4FeS + 12NH 4 C1 + S 2 

Fe 2 S 3 + 6HC1 = 3H 2 S + 2 F 6 CI, 

2FeCl 3 + 3Na 2 CO s + 3H 2 0 = 2 Fe(OH ) 3 + 6 NaCl + 3C0 2 
FeCl 3 + NasHPC^ = FeP0 4 + 2NaCl + HC1 
FeCl 3 + 3 NH 4 CNS = Fe(CNS ) 3 + 3NH 4 C1 
4FeCl 3 + 3 K 4 Fe(CN ) 6 = 12KC1 + Fe 4 (Fe(CN ) 6 ) 3 
FeCl 3 + K 3 Fe(CN ) 6 = 3KC1 + Fe(Fe(CN) 6 ) 

CHROMIUM SALTS 

Cr 2 (S0 4 ) 3 + 6 NH 4 OH = 2Cr(OH ) 3 + 3 (NH 4 ) 2 S 04 

Cr 2 (S0 4 ) 3 + NH 4 CI + 6 NH 4 OH = 2Cr(OH ) 3 + 3(NH 4 ) 2 S0 4 + NH 4 C1 

Cr(OH ) 3 + NH 4 CI + NH 4 OH = no reaction 

Cr 2 (S0 4 ) 3 + 6 NaOH = 2Cr(OH ) 3 + 3 Na 2 S 04 (Cr(OH ) 3 + NaOH = NaCr0 2 + 2H a O) 

Cr(ONa ) 3 + 3H 2 0 + heat = Cr(OH ) 3 + 3NaOH 

Cr 2 (S0 4 ) 3 + 3 (NH 4 ) 2 S + 6H 2 0 = 2Cr(OH ) 3 + 3(NH 4 ) 2 S0 4 + 3H 2 S 

Cr 2 (S0 4 ) 3 + 3Na. 2 C0 3 + 3H 2 0 = 2 Cr(OH ) 3 + 3Na 2 S0 4 + 3C0 2 

Cr 2 (S0 4 ) 3 + 2Na 2 HP0 4 = CrP0 4 + Na 2 S0 4 + 2NaHS0 4 

CrCl 3 + 3 NH 4 OH = Cr(OH ) 3 + 3NH 4 C1 

Cr(NH 3 ) 4 Cl 3 + 4H 2 0 + heat = Cr(OH ) 3 + 3NH 4 C1 + NH 4 OH 

CrCl 3 + NH 4 CI + 3 NH 4 OH = Cr(OH ) 3 + 3NH 4 C1 

CrCl 3 + 3 NaOH = Cr(OH ) 3 + 3NaCl (Cr(OH ) 3 + NaOH = NaCr0 2 + 2H 2 0) 

2CrCl 3 + 3 (NH 4 ) 2 S + 6H 2 0 = 2 Cr(OH ) 3 + 6 NH 4 CI + 3H 2 S 
2CrCl 3 + 3Na 2 C0 3 + 3H 2 0 = 2 Cr(OH ) 3 + 6 NaCl + 3C0 2 
CrCl 3 + Na-,HP0 4 = CrP0 4 + 2 NaCl + HC1 

Cr 2 (S0 4 ) 3 + 3KN0 3 + 2 Na 2 C 0 3 - 2Na 2 Cr0 4 + 3KN0 2 + 2C0 2 + 3S0 2 
Cr(ONa ) 3 + 2H 2 0 2 = Na 2 Cr0 4 + 2H 2 0 

2Cr(OH ) 3 + 3KN0 3 + 2Na 2 C0 3 = 2Na 2 Cr0 4 + 3KN0 2 + 2C0 2 + 3H 2 0 
2K 2 Cr 2 0 7 + 16HC1 + 6 H 2 S = 4KC1 + 4CrCl 3 + 14H 2 0 + 6S 
Na 2 Cr0 4 + (CH 3 COO) 2 Pb = PbCr0 4 + 2 CH 3 COONa 
Cr(OH ) 3 + 3HC1 = CrCl 3 + 3H 2 0 
Cr(OH ) 3 + 3HNO s = Cr(N0 3 ) 3 + 3H 2 0 

K 2 Cr 2 0 7 + 7H 2 S0 4 + 3 Zn = K 2 S0 4 + Cr 2 (S0 4 ) 3 + 3ZnS0 4 + 7H 2 0 

183 


ALUMINUM SALTS 


A1 2 (S0 4 ) 3 + 6 NH 4 OH = 2A1(0H ) 3 + 3(NH 4 ) 2 S0 4 

A1 2 (S0 4 ) 3 + NH 4 C1 + 6 NH 4 OH = 2A1(0H ) 3 + 3(NH 4 ) 2 S0 4 + NH 4 C1 

A1 2 (S0 4 ) 3 + 6 NaOH = 2A1(0H ) 3 + 3Na 2 S0 4 

Al(OH), + 3 NaOH = Na 3 A10 3 + 3H 2 0 

A1 2 (S0 4 ) 3 + 3 (NH 4 ) 2 S + 6H 2 0 = 2A1(0H ) 3 + 3(NH 4 ) 2 S0 4 + 3H 2 S 

A1 2 (S0 4 ) 3 + 3Na 2 C0 3 + 3H 2 0 = 2A1(0H ) 3 + 3Na 2 S0 4 + 3C0 2 

A1 2 (S0 4 ) 3 + 2Na 2 HP0 4 = 2A1P0 4 + Na*S0 4 + 2NaHS0 4 

AIC1 3 + 3NH 4 OH = Al(OH), + 3NH 4 C1 

A1C1 3 + 3NaOH = Al(OH). + 3NaCl 

2A1C1 3 + 3 (NH 4 ) 2 S + 6H 2 0 = 2A1(0H ) 3 + 6NH 4 C1 + 3H 2 S 

2A1C1 3 + 3Na 2 C0 3 + 3H 2 0 = 2A1(0H ) 3 + 6 NaCl + 3C0 2 

Al(OH), + 3HC1 = A1C1 3 + 3H 2 0 

Al(OH), + 3HN0 3 = A1(N0 3 ) 3 + 3H 2 0 

NICKEL SALTS 

Ni(N0 3 ) 2 + HC1 +H 2 S = no action 

Ni(N0 3 ) 2 + 2 NH 4 OH = Ni(OH ) 2 + 2NH 4 N0 3 

Ni(N0 3 ) 2 + 2NH 4 C1 + 2NH 4 OH = Ni(NH 3 ) 4 Cl 2 + 4H 2 0 

Ni(N0 3 ) 2 + NH 4 C1 + NH 4 OH + (NH 4 ) 2 S = NiS + 2NH 4 N0 3 

Ni(N0 3 ) 2 + 2 NaOH = Ni(OH ) 2 + 2NaN0 3 

Ni(OH ) 2 + heat = NiO + H 2 0 

Ni(N0 3 ) 2 + Na 2 C 0 3 = NiC0 3 + 2NaN0 3 

Ni(N0 3 ) 2 + CH 3 (C(NOH)) 2 CH 3 = red ppt. 

Ni(N0 3 ) 2 + Na 2 B 4 0 7 = fusion 

Ni(NH 3 ) 4 (N0 3 ) 2 + H 2 S = Ni(N0 3 ) 2 + 2(NH 4 ) 2 S + S 

Ni(N0 3 ) 2 + (NH 4 ) 2 S = NiS + 2NH 4 N0 3 

NiS + HC1 = no action. 

3NiS + 8HN0 3 = 3Ni(N0 3 ) 2 + 4H 2 0 + 2NO + 3S 
3 NiS + 6HC1 + 2HN0 3 = 3NiCl 2 + 4H 2 0 + 2NO + 3S 
NiS + Na^B^? = fusion. 

NiCl 2 + HC1 + H 2 S = no action 

NiCl 2 + NH 4 C1 + NH 4 OH + (NH 4 ) 2 S = NiS + 2NH 4 C1 
NiCl 2 + 2NaOH = Ni(OH ) 2 + 2NaCl 
NiCl 2 4- Na 2 C0 3 = NiC0 3 + 2NaCl 


COBALT SALTS 

Co(N0 3 ) 2 + HC1 + H 2 S = no action 
Co(N0 3 ) 2 + 2 NH 4 OH = Co(OH) 2 + 2NH 4 N0 3 
Co(N0 3 ) 2 + NH 4 C1 + 2NH 4 OH = Co(OH) 2 + 2NH 4 N0 3 
Co(N0 3 ) 2 + NH 4 C1 + NH 4 OH + (NH 4 ) 2 S = CoS + 2NH 4 N0 3 

184 


Co(N0 3 ) 2 + 2NaOH = Co(OH ) 2 + 2NaN0 3 
Co(OH) 2 + heat = Co + H 2 0 
Co(N0 3 ) 2 + Na 2 C0 3 = blue color 

Co(N0 3 ) 2 + HC1 + 2C 10 H 6 NO(OH) = Co(C 10 H 6 (NO)O ) 2 + 2HN0 3 
Co(N0 3 ) 2 + Na 2 B 4 07 = blue color 
CoS + HC1 = no action 

3CoS 8HN0 3 = 4H 2 0 -f- 2NO 3S -f~ 3Co(N0 3 ) 2 

3CoS + 6HC1 + 2HN0 3 = 3CoC1 2 + 2NO + 4H 2 0 + 3S 

CoS + Na 2 B 4 0 7 = blue color 

CoCl 2 + HC1 + H 2 S = no action 

CoCl 2 + 2 NH 4 OH = Co(OH) 2 + 2 NH 4 CI 

C 0 CI 2 + NH 4 CI + 2 NH 4 OH = Co(OH) 2 + 2 NH 4 CI 

CoCl 2 + NH4CI + NH 4 OH + (NH 4 ) 2 S = CoS + 2 NH 4 CI 

CoCl 2 + 2NaOH = Co(OH ) 2 + 2NaCl 

MANGANESE SALTS 

MnS0 4 + HC1 + H 2 S = no action 

MnS0 4 + 2 NH 4 OH = Mn(OH ) 2 + (NH 4 ) 2 S0 4 

MnS0 4 + NH4CI + 2 NH 4 OH = Mn(OH ) 2 + (NH 4 ) 2 S0 4 

MnS0 4 + NH4CI + NH4OH + (NH 4 ) 2 S = MnS + (NH 4 ) 2 SO 

MnS0 4 + 2 NaOH = Mn(OH ) 2 + Na 2 S0 4 

Mn(OH ) 2 + 0 2 = H 2 Mn0 4 

2Mn(OH ) 2 = Mn 2 0 3 T H 2 0 

MnS0 4 T 2Na 2 C0 3 -t - 0 2 == 2C0 2 -f- Na 2 S0 4 T Na 2 Mn0 4 
4MnC0 3 + 30 2 + 2H 2 0 = 4HMn0 3 + 3C0 2 

2MnS0 4 + 5Pb0 2 + 6HN0 3 = 2PbS0 4 + Pb(N0 3 ) 2 + 2HMn0 4 + 2H 2 0 
2MnS0 4 + 5Pb 2 0 3 + 16HN0 3 = 2HMn0 4 + 2PbS0 4 + 8Pb(N0 3 ) 2 + 7H 2 0 
MnS0 4 + Na 2 B 4 0 7 = fusion. 

MnS0 4 + KN0 3 + Na 2 C0 2 = KN0 2 + C0 2 + Na 2 Mn0 4 

MnS0 4 + (NH 4 ) 2 S = MnS + (NH 4 ) 2 S0 4 

MnS + 2HC1 = MnCl 2 + H 2 S 

3MnS + 8HN0 3 = 3Mn(N0 3 ) 2 + 4H 2 0 + 2NO + 3S 

2KMn0 4 + 10FeSO 4 + 9H 2 S0 4 = 5Fe 2 (S0 4 ) 3 + 2MnS0 4 + 2KHS0 4 + 8H 2 0 

Mn0 2 + H 2 S0 4 + H 2 C 2 0 4 = MnS0 4 + 2H 2 0 + 2C0 2 

3Mn0 2 + 6 KOH + KC10 3 = 3K 2 Mn0 4 + KC1 + 3H 2 0 

2KMn0 4 + 10KI + 8H 2 S0 4 = 6K 2 S0 4 + 2MnS0 4 + 8H 2 0 + 5I 2 

ZINC SALTS 

ZnS0 4 + HC1 + H 2 S = no action 

ZnS0 4 + 2 NH 4 OH = Zn(OH ) 2 + (NH 4 ) 2 S0 4 

ZnS0 4 + NH4CI + NH4OH = no action 

ZnS0 4 + NH4CI + NH 4 OH + (NH 4 ) 2 S = ZnS + (NH 4 ) 2 S0 4 


185 


Zn(NH 3 ) 4 S0 4 4 (NH 4 ) 2 S + H 2 0 = ZnS 4 (NH 4 ) 2 S0 4 4 4NH 4 OH 
ZnS0 4 + 2NaOH = Zn(OH ) 2 + Na 2 S0 4 

Zn(OH ) 2 + 2 NaOH = Na 2 Zn0 2 + 2H 2 0 or ZnS0 4 + 4NaOH = Na 2 Zn0 2 4 2H 2 0 4 Na 2 S0 4 

5ZnS0 4 4 5Na 2 C0 3 + 3H 2 0 = Zn 5 (0H)(C0 3 ) 2 + 5N & 2 S0 4 4 3C0 2 

ZnS0 4 4 H 2 S = ZnS 4 H£Ot 

ZnS + 2HC1 = ZnCl 2 4 H 2 S 

3ZnS + 8 HNO 3 = 3Zn(N0 3 ) 2 + 4H 2 0 4 2 NO 

Zn(NH 3 ) 4 S0 4 4 H 2 S = ZnS + (NH 4 ) 2 S0 4 4 (NH 4 ) 2 S 

Na 2 Zn0 2 + 4CH 3 COOH = (CH 3 COO) 2 Zn 4 2 CH 3 COONa + 2H 2 0 

ZnCl 2 + NH 4 OH = Zn(OH ) 2 + 2NH 4 C1 

ZnCl 2 + NH 4 C1 4 NH 4 OH + (NH 4 ) 2 S = ZnS + 2NH 4 C1 

ZnCl 2 + NaOH = Zn(OH ) 2 + 2NaCl or 4NaOH + ZnCl 2 = Na 2 Zn0 2 4 2H 2 0 4 2NaCl 

ZnCl 2 4 Na 2 C0 3 + 2H 2 0 = Zn(OH ) 2 + 2NaCl + C0 2 4 H 2 0 

ZnCl 2 + H 2 S = ZnS + 2HC1 

Zn(NH 3 ) 4 Cl 2 + 2H 2 S = ZnS 4 2NH 4 C1 4 (NH 4 ) 2 S 

CALCIUM SALTS 

CaCl 2 4 HC1 4 H 2 S 4 (NH 4 ) 2 S = no precipitate 

CaCl 2 4 NH 4 C1 4 NH 4 OH 4 (NH 4 ) 2 C0 3 = CaCl 3 4 3NH 4 C1 4 NH 4 OH 

CaCl 2 4 K 2 Cr 2 0 7 = no ppt. distinction from Ba 

CaCl 2 4 (COONH 4 ) 2 = Ca(COO ) 2 4 2NH 4 C1 

(COO) 2 Ca 4 CH 3 COOH = insoluble 

CaCl 2 4 H 2 S0 4 = CaS0 4 4 2HC1 (from con. solutions) 

CaCl 2 4 CaS0 4 = no action 

CaCl 2 4 (NH 4 ) 2 S0 4 = CaS0 4 4 2NH 4 C1 (sol. in excess hot (NH 4 ) 2 S0 4 ) 

CaCl 2 4 Na 2 HP0 4 = CaHP0 4 4 2NaCl 

CaCl 2 4 Na 2 C0 3 = CaC0 3 4 2NaCl 

CaCl 2 4 2NaOH = Ca(OH ) 2 4 2NaCl (con. sol.) 

(COO) 2 Ca 4 2HC1 = CaCl 2 4 H 2 0 4 C0 2 4 CO 

MAGNESIUM SALTS 

MgS0 4 4 HC1 4 H 2 S 4 NH 4 C1 4 NH 4 OH 4 (NH 4 ) 2 S 4 (NH 4 ) 2 C0 3 = no precipitate. 
MgS0 4 4 NH 4 C1 4 NH 4 OH 4 (COONH 4 ) 2 = no ppt. 

2MgS0 4 4 2NH 4 OH = Mg(OH ) 2 4 (NH 4 ) 2 Mg(S0 4 ) 2 (precipitation not comp, due to sol. in 
NH 4 OH sol.) 

Mg(OH ) 2 4 2NH 4 C1 = NH 4 MgCl 3 4 2 NH 4 OH 
MgS0 4 4 2 NaOH = Mg(OH), 4 Na^SO, 

MgS0 4 4 Na 2 C0 3 = MgC0 3 4 Na 2 S0 4 
MgC0 3 4 3NILC1 = NH 4 MgCl 3 4 (NH 4 ) 2 C0 3 
MgS0 4 4 (NH 4 ) 2 S0 4 = no action 

MgS0 4 4 NH 4 C1 4 NH 4 OH 4 Na 2 HP0 4 = MgNH 4 P0 4 4 Na 2 S0 4 4 NH 4 C1 4 H 2 0 
MgNH 4 P0 4 4 2HC1 = MgCl 2 4 NH 4 H 2 P0 4 

186 * 


MgNH 4 P0 4 + 2HN0 3 = Mg(N0 3 ) 2 + NH 4 H 2 P0 4 
MgNH 4 P0 4 + H 2 S0 4 = MgS0 4 + NH 4 H 2 P0 4 
MgNH 4 P0 4 + 2CH 3 COOH = Mg(CH 3 COO) 2 + NH 4 H 2 P0 4 
MgNH 4 P0 4 + NH 4 OH = no action 
MgNH 4 P0 4 + 2NaOH = Mg(OH) 2 + NH 4 Na 2 P0 4 
MgS0 4 + 2NH 4 OH = (NH 4 ) 2 S0 4 + Mg(OH) 2 
Mg(OH) 2 + 2NH 4 C1 = MgCl 2 + 2NH 4 OH 

STRONTIUM SALTS 

SrCl 2 + NH 4 C1 + NH 4 OII + (NH 4 ) 2 C0 3 = SrC0 3 + 2NH 4 C1 

SrCr0 4 + 2CH 3 COOH = Sr(C 2 H 3 0 2 ) 2 + H 2 Cr0 4 

SrCl 2 + (COONH 4 ) 2 = (COO ) 2 Sr + 2NH 4 C1 

(COO)aSr + 2CH 3 COOH = Sr(C 2 H 3 0 2 ) 2 + (COOH) 2 

SrCl 2 + H 2 S0 4 = SrS0 4 + 2HC1 

SrS0 4 + 2HC1 = SrCl 2 + H 2 S0 4 (boiling) 

SrCl 2 + CaS0 4 = SrS0 4 + CaCl 2 
SrCl 2 + (NH 4 ) 2 S0 4 = SrS0 4 + 2NH 4 C1 
SrCl 2 + Na 2 HP0 4 = SrHP0 4 + 2NaCl 
SrCl 2 + Na 2 C0 3 = SrC0 3 + 2NaCl 
SrHP0 4 + 2HC1 = SrCl 2 + H 3 P0 4 
(COO) 2 Sr + 2HN0 3 = (COOH) 2 + Sr(N0 3 ) 2 
(COO) 2 Sr + 2HC1 = (COOH) 2 + SrCl 2 

BARIUM SALTS 

BaCl 2 + NH 4 C1 + NH 4 OH + (NH 4 ) 2 C0 3 = BaC0 3 + NH 4 C1 

2BaCl 2 + K 2 Cr 2 0 7 + H 2 0 2 = 2BaCr0 4 + 2KC1 + 2HC1 

BaCr0 4 + 2CH 3 COOH = no action 

BaCl 2 + (COOH) 2 = BaC 2 0 4 + 2HC1 

(COO) 2 Ba + 2CH 3 COOH = Ba(C 2 H 3 0 2 ) 2 + (COOH) 2 

BaCl 2 + H 2 S0 4 = BaS0 4 + 2HC1 

BaCl 2 + CaS0 4 = BaS0 4 + CaCl 2 

BaCl 2 + (NH 4 ) 2 S0 4 = BaS0 4 + 2NH 4 C1 

BaCl 2 + Na 2 HP0 4 = BaHP0 4 + 2NaCl 

BaCl 2 + Na 2 C0 3 = BaC0 3 + 2NaCl 

BaC0 3 + 2HC1 = BaCl 2 + H 2 0 + C0 2 

BaHP0 4 + 2HC1 = BaCl 2 + H 3 P0 4 

(COO) 2 Ba + 2HN0 3 = Ba(N0 3 ) 2 + (COOH) 2 

SODIUM SALTS 

NaCl + Na 3 Co(N0 2 ) 6 = no action 

NaCl + Na 3 Co(N0 2 ) c + CH 3 COOH = no action 


187 


POTASSIUM SALTS 

3KC1 + Na 3 Co(N0 2 ) 6 = K 3 Co(N0 2 ) 6 + 3NaCl 

3KC1 + Na 3 Co(N0 2 ) 6 + CH 3 COOH = K 3 Co(N0 2 ) 6 + 3NaCl + CH 3 COOH 
2KC1 + Na 3 Co(N0 2 ) 6 = K 2 Na(Co(N0 2 ) 6 ) + 2NaCl 

AMMONIUM SALTS 

NH 4 CI + NaOH = NH 3 + NaCl + H 2 0 
NH 4 CI + KOH = NH 3 + KC1 + H 2 0 
NH 4 CI + CaS0 4 = no action 

3NH 4 C1 + Na 3 Co(N0 2 ) 6 = 3 NaCl + (NH4) 3 Co(N0 2 ) 6 

NH 4 CI solid heated = NH 3 + HC1 

NH 4 N0 2 solid heated = 2H 2 0 + N 2 

(NH 4 ) 2 S 04 solid heated = sublimes 

2 NH 4 CI + CaO solid heated = CaCl 2 + H 2 0 + 2NH 3 


188 




SUMMARIZED GROUP REACTIONS 


FIRST GROUP 

LEAD 

Pb(N0 3 ) 2 + 2HC1 = PbCl 2 + 2HN0 3 

white ppt.' 

PbCl 2 + hot H 2 0 = Solution 
PbCl 2 + K 2 Cr0 4 = PbCr0 4 + 2KC1 

yellow ppt. 

SILVER 

AgN0 3 + HC1 = AgCl + HN0 3 

white ppt. 

AgCl + 2NH 4 OH = Ag(NH 3 ) 2 Cl + 2H 2 0 
Ag(NH 3 ) 2 Cl + 2HN0 3 = AgCl + 2NH 4 N0 3 

white ppt. 

MERCURY (OUS) 

HgN0 3 + HC1 = HgCl + HN0 3 

white ppt. 

2HgCl + 2NH 4 OH = HgNHoCl.Hg + NH 4 C1 + 2H 2 0 

black residue on filter 


SECOND GROUP —A 

MERCURY (IC) 

HgCl 2 + H 2 S = HgS + 2HC1 

black ppt. HgS + (NHihSx = insoluble 

HgS + HN0 3 = insoluble 

3HgS + 2HN0 3 + 6HC1 = 3HgCl 2 + 2NO + 4H 2 0 + 3S 
HgCl 2 + SnCl 2 = SnCl 4 + Hg 

black ppt. 

LEAD 

PbCl 2 + H 2 S = PbS + 2HC1 

black ppt. PbS + (NHihSx = insoluble 

3PbS + 8HN0 3 = 3Pb(N0 3 ) 2 + 2NO + 4H 2 0 + 3S 
Pb(N0 3 ) 2 + H 2 S0 4 = PbS0 4 + 2HN0 2 

white ppt. 

189 


BISMUTH 


2BiCl 3 + 3H 2 S = Bi 2 S 3 + 6HC1 

brown ppt. 

Bi 2 S 3 + (NH 4 ) 2 Sx = insoluble 
Bi 2 S 3 + 8HN0 3 = 2Bi(N0 3 ) 3 + 2NO + 4H 2 0 + 3S 
2Bi(N0 3 ) 3 + 3H 2 S0 4 = Bi 2 (S0 4 ) 3 + 6HNO s 
Bi 2 (S0 4 ) 3 + 6NH 4 OH = 2Bi(OH) 3 + 3(NH 4 ) 2 S0 4 

white ppt. 

f Bi(OH) 3 + 3HC1 = BiCl 3 + 3H 2 0 
\ BiCl 3 + H 2 0 = BiOCl + 2HC1 

white ppt. 
or 

2Bi(OH) 3 + 3Na 2 Sn0 2 = 2 Bi + 3Na 2 Sn0 3 + 3H 2 0 

black residue on filter 

COPPER 

CuCl 2 + H 2 S = CuS + 2HC1 

black ppt. 

3CuS + 8HN0 3 = 3 Cu(N0 3 ) 2 + 2NO + 4H 2 0 + 3S 
Cu(N 0 3 ) 2 + H 2 S0 4 = CuS0 4 + 2HN0 3 
CuS0 4 + 4NH 4 OH = Cu(NH 3 ) 4 S0 4 + 4H 2 0 

deep blue coloration 

or 

8H 2 0 + 2Cu(NH 3 ) 4 S0 4 + K 4 Fe(CN) 6 = Cu 2 Fe(CN) 6 + 2K 2 S0 4 + 8NH 4 OH 

brick red ppt. 

CADMIUM 

CdCl 2 + H 2 S = CdS + 2HC1 

yellow ppt. 

3CdS + 8HN0 3 = 3Cd(N0 3 ) 2 + 2NO + 4H 2 0 + 3S 

Cd(N0 3 ) 2 + H 2 S0 4 = CdS0 4 + 2HN0 3 

CdS0 4 + 4NH 4 OH = Cd(NH 3 ) 4 S0 4 + 4H 2 0 

Cd(NH 3 ) 4 S0 4 + (NH 4 ) 2 S + 4H 2 0 = (NH 4 ) 2 S0 4 + 4NH 4 OH + CdS 

yellow ppt. 


SECOND GROUP —B 


ARSENIC (OUS) 

2Na 3 As0 3 + 3H 2 S + 0HC1 = 6NaCl + As 2 S 3 + 6H 2 0 

yellow ppt. 

As 2 S 3 + (NH 4 ) 2 Sx = 2NH 4 AsS 2 + (x-l)S 
2NH 4 AsS 2 + 2HC1 = As 2 S 3 + 2NH 4 C1 + H 2 S 

yellow ppt. 


3As 2 S 3 + 10HNO 3 + 4H 2 0 =. 6H 3 As0 4 + 10NO + 9S 

N 3 As0 4 + MgS0 4 + NH 4 C1 + 3NH 4 OH = MgNH,As0 4 + (NH 4 ) 2 S0 4 +NH 4 C1+3H 2 0 

white crystalline ppt. 

190 


ARSENIC (IC) 


2Na 3 As0 4 -f- 5 H 2 S -f- 6HC1 = AsoSs + 6NaCl + 8 H 2 O 
As 2 S 6 + 3(NH 4 ) 2 Sx = 2(NH 4 )sAsS 4 + (3x - 3)S 
2(NH 4 ) 3 AsS 4 + 6HC1 = 6NH 4 C1 + As 2 S 5 + 3H 2 S 

yellow ppt. 

3 AS 2 S 5 IOHNO3 + 4H2O = 6HjAs0 4 -f- IONO -j* 15S 
Final test same as above 


ANTIMONY 

2SbCl 3 + 3H 2 S = Sb 2 S 3 + 6HC1 

orange ppt. 

Sb 2 S 3 + 3(NH 4 ) 2 Sx = 2(NH 4 ) 3 SbS 4 + (3* - 5)S 


| antimonous 


2SbCl s + 5H 2 S = Sb 2 S 5 + 10HC1 

orange ppt. 

Sb 2 S 5 + 3(NH 4 ) 2 Sx = 2(NH 4 ) 3 SbS 4 + (3a; - 3)S 


dilute 

2(NH 4 ) 3 SbS 4 + 6HC1 = Sb 2 S 6 + 6NH 4 C1 + 3H 2 S 

orange ppt. 


> antimonic 


cone. 

Sb 2 S 6 + 10HC1 = 2SbCl 6 + 5H 2 S 
2SbCl s + 5H 2 S = Sb 2 S 6 + 10 HCL 

in dilute sol. orange ppt. 


TIN 

SnCl 2 + H 2 S = SnS + 2HC1 

dark brown 


SnS + (NH 4 ) 2 Sx = (NH 4 ) 2 SnS 2 + {x - 2)S 
SnCl 4 + 2H 2 S = SnS 2 + 4HC1 

yellow ppt. 

SnS 2 + (NH 4 ) 2 Sx = (NH 4 ) 2 SnS 3 + (* - 1)S 

dilute 

(NH 4 ) 2 SnS 3 + 2HC1 = 2NH 4 C1 + SnS 2 + H 2 S 

yellow ppt. 


cone. 

SnS 2 + 4HC1 = SnCl 4 + 2H 2 S 
SnCU + 2H 2 S = SnS 2 + 4HC1 

very dilute sol. yellow ppt. 

or SnCl 4 + Zn + 2HC1 = SnCl 2 + ZnCl 2 + 2HC1 
SnCl 2 + HgCl 2 = SnCl 4 + Hg 

black ppt. 


191 






THIRD GROUP 


IRON 

2FeCl 3 + H 2 S = 2FeCl 2 + 2HCI + S 

6FeCl 2 + 8 HNO 3 = 4FeCl 3 + 2Fe(N0 3 ) 3 + 2 NO + 4H 2 0 

FeCl 3 + 3NH 4 OH = Fe(OH) 3 + 3NH 4 C1 

reddish brown ppt. 

Fe(N0 3 ) 3 + 3NH 4 OH = Fe(OH) 3 + 3NH 4 N0 3 

reddish brown ppt. 

Fe(OH) 3 + 3HCI = FeCl 3 + 3H 2 0 
FeCl 3 + 3NH 4 CNS = Fe(CNS) 3 + 3NH 4 C1 

red coloration 

4FeCl 3 + 3K 4 Fe(CN) 6 = 12 KC1 + Fe 4 (Fe(CN) 6 ) 3 

deep blue ppt. 

ALUMINUM 

A1CU + 3NH 4 OH = 3NH 4 C1 + Al(OH), 

white gelatinous ppt. 

Al(OH) 3 + 3NaOH = Na 3 A10 3 + 3H 2 0 
Na 3 A10 3 + 6HC1 = 3NaCl + A1C1* + 3H 2 0 
A1CU + 3NH 4 OH = 3NH 4 C1 + Al(OH) 3 

white gelatinous ppt. 

2A1(0H) 3 + 3CoC 1 2 = Co 3 (A10 3 ) 2 + 6HC1 

blue coloration 

N.B. Cr and Zn interfere with this test. 

CHROMIUM 

hot 

CrCl 3 + 3NH 4 OH = Cr(OH) 3 + 3NH 4 C1 

light green gelatinous ppt. 

2Cr(OH) 3 + 2Na 2 C0 3 + 3KN0 3 = 3KN0 2 + 2C0 2 + 3H 2 0 + 2Na 2 Cr0 4 

.. yellow fusion 

Na 2 Cr0 4 + Pb(CH 3 COO) 2 = 2CH 3 COONa + PbCr 04 

yellow ppt. 


FOURTH GROUP 

COBALT 

CoCl 2 + (NH 4 ) 2 S = CoS + 2 NH 4 CI 

black ppt. 

dilute 

CoS + HC1 = insoluble 

2CoS + 2Na 2 B 4 0 7 + 30 2 = 2Co(B0 2 ) 2 .2NaB0 2 + 4S0 2 

blue bead 

NICKEL 

NiCl 2 + (NH 4 ) 2 S = NiS + 2 NH 4 CI 

black ppt. 

NiS + HC1 = insoluble 

3NiS + 6HC1 + 2HN0 3 = NiCl 2 + 2NO + 4H 2 0 + 3S 

2(CH 3 ) 2 C 2 N 2 0 2 H 2 + NiCl 2 + 2 NH 4 OH = 2 NH 4 CI + 2H 2 0 + ((CH 3 ) 2 C 2 N 2 0 2 )Ni 

bright red ppt. 


192 



MANGANESE 


MnCl 2 + (NH 4 ) 2 S = 2NH 4 C1 + MnS 

flesh-colored ppt. 

dilute 

MnS + 2HC1 = MnCl 2 + H 2 S 

2MnCl 2 + 5Pb0 2 + 6 HNO 3 = 2HMn0 4 + 2PbCl 2 + 3Pb(N0 3 ) 2 + 2H 2 0 

purple filtrate 


ZINC 

ZnCl 2 + (NH 4 ) 2 S = ZnS + 2NH 4 C1 

white ppt. 

dilute 

ZnS + 2HC1 = ZnCl 2 + H 2 S 

ZnCl 2 + 4NaOH = Na 2 Zn0 2 + 2NaCl + 2H 2 0 

no ppt. 

Zn(CH 3 COO ) 2 + H 2 S = ZnS + 2CH 3 COOH 

white ppt. 

N.B. ZnS is best precipitated from solution slightly acid with CH 3 COOH to which NH 4 CH 3 COO 
has been added to reduce the hydrogen ion concentration. 


FIFTH GROUP 

BARIUM 

BaCl 2 + (NH 4 ) 2 C0 3 = BaC0 3 + 2NH 4 C1 

white ppt. 

BaC0 3 + 2CH 3 COOH = Ba(CH 3 COO) 2 + II 2 0 + C0 2 

2Ba(CH 3 COO) 2 + K 2 Cr 2 0 7 + H 2 0 = 2BaCr0 4 + 2CH 3 COOH + 2CH 3 COOK 

white ppt. 

m 

BaCr0 4 + HC1 = green flame 

CALCIUM 

CaCl 2 + (NH 4 ) 2 C0 3 = CaC0 3 + 2NH 4 C1 

white ppt. 

CaC0 3 + 2CH 3 COOH = Ca(CII 3 COO) 2 + H 2 0 + C0 2 
Ca(CH 3 COO) 2 + CH 3 COOH + K 2 Cr 2 0 7 = reversible reaction 

no ppt. 

Ca(CH 3 COO) 2 + (NH 4 ) 2 C0 3 = CaC0 3 + 2(NH 4 )CH 3 COO 

white ppt. 

CaC0 3 + 2HC1 = CaCl 2 + H 2 0 (- C0 2 
CaCl 2 + (NH 4 ) 2 S0 4 = CaS0 4 + 2NH 4 C1 

only partial precipitation, as CaS0 4 is somewhat soluble 

CaS0 4 + (NH 4 ) 2 C 2 0 4 = CaC 2 0 4 + (NH 4 ) 2 S0 4 

white ppt. 

CaC 2 0 4 + HC1 = brick red for second, then yellow flame 

193 


STRONTIUM 


SrCI 2 + (NH 4 ) 2 C0 3 = SrC0 3 + 2NH 4 C1 

white ppt. 

SrC0 3 + 2CH 3 COOH = Sr(CH 3 COO) 2 + H 2 0 + C0 2 
Sr(CH 3 COO) 2 + CH 3 COOH + Iv 2 Cr 2 07 = reversible reaction 

no ppt. 

Sr(CH 3 COO) 2 + (NH 4 ) 2 C0 3 = SrC0 3 + 2NH 4 CH 3 COO 

white ppt. 

SrC0 3 + 2HC1 = SrCl 2 + H 2 0 + C0 2 
SrCl 2 + CaS0 4 = SrS0 4 + CaCl 2 

white ppt. 

SrCl 2 + (NH 4 ) 2 S0 4 = SrS0 4 + 2 NH 4 CI 

white ppt. 

SrS0 4 + HC1 = red flame 

MAGNESIUM 

MgCl 2 + Na 2 HP0 4 + NH 4 OH = 2NaCl + H 2 0 + MgNH 4 P0 4 

white crystalline ppt. 

AMMONIUM 

NH 4 C1 + NaOH = NaCl + NH 3 + H 2 0 

SODIUM 

NaCl + HC1 = yellow flame 

POTASSIUM 

KC1 + HC1 = violet flame 

3KC1 + Na 3 Co(N0 2 )e = K 3 Co(N0 2 ) 6 + 3NaCl 

yellow ppt. 

N.B. Ammonium salts give the same test with sodium; cobaltic nitrite, therefore, must be removed 
before making this test. 


194 


ACID TESTS “GROUP A” 


Preliminary test in presence of dilute HNO ; 
M 2 S + 2AgN0 3 = Ag 2 S + 2MN0 3 

black ppt. 


MCI + AgN0 3 = AgCl + MN0 3 

white ppt. 

MBr + AgN0 3 = AgBr -f- MN0 3 

cream-colored ppt. 

MI + AgN0 3 = Agl + MN0 3 

yellow ppt. 

M*S + CdS0 4 = CdS + M 2 S0 4 

yellow ppt. 

MCI + AgN0 3 = AgCl + MN0 3 

white ppt. 


SULFIDES 

CHLORIDES 


AgCl + 2NH 4 OH = Ag(NH 3 ) 2 Cl + 2H 2 0 

BROMIDES 

2MBr + Cl 2 = 2MC1 + Br 2 

Br 2 + CS 2 = reddish brown solution 

IODIDES 

2MI + Cl 2 = 2MC1 + I 2 
I 2 + CS 2 = purple solution 

ACID TESTS “GROUP B ” 

Preliminary tests in neutral solution. 

M 2 C0 3 + BaCl 2 = BaC0 3 + 2MC1 

white ppt. 

2M 3 P0 4 + 3BaCl 2 = Ba 3 (P0 4 ) 2 + 6MC1 

white ppt. 


M 2 S0 4 + BaCl 2 = BaS0 4 + 2MC1 

white ppt. 

M 2 S0 4 + BaCl 2 = BaS0 4 + 2MC1 

white ppt. 

BaS0 4 + HC1 = insoluble 
M 2 C0 3 + BaCl 2 = BaC0 3 + 2MC1 

white ppt. 


SULFATES 


CARBONATES 


BaC0 3 + 2HC1 = BaCl 2 + H 2 0 + C0 2 
C0 2 + Ca(OH) 2 = CaC0 3 + H 2 0 

white ppt. 


PHOSPHATES 

M 3 P0 4 + 12(NH 4 ) 2 Mo0 4 + 2HN0 3 = 21NH 4 N0 3 + 3MN0 3 + 12H 2 0 + (NH 4 ) 3 P0 4 .12Mo0 3 

yellow pp 

Arsenic gives a similar precipitate, hence must be removed before testing for phosphates. 

195 


“GROUP C” 

No preliminary test. 

NITRATE 

2MN0 3 + 4H 2 S0 4 + 10FeS0 4 = M2SO4 + 3Fe 2 (S0 4 ) 3 + 4H 2 0 + 2(FeS0 4 ) ? N0 

brown ring 

N. B. The addition of FeS0 4 to the unknown and then pouring concentrated H 2 S0 4 carefully 
down the side of the tube gives the best results. 

NOTES 

(1) Before testing for Group B acids, Group I metals should be removed, otherwise a white ppt., 
insoluble in HC1, will always be obtained. 

Thus: 2 AgN 03 + BaCl 2 = 2 AgCl + Ba(N 0 3 )2 

white ppt. in HC1 

(2) Never use any filtrate other than that obtained from the precipitation of Group I metals 
for testing for acid radicals other than phosphate, for under the proper conditions the following 
reactions may take place: 

NaN0 3 + HC1 = NaCl + HN0 3 
2HN0 3 + 3H 2 S = 4H 2 0 + 2 NO + 3S 
or 

2HN0 3 + H 2 S = H 2 SO 3 + 2 NO + H 2 0 
or 

8 HNO 3 + 3H 2 S = 3H 2 S0 4 + 8 NO + 4H 2 0 

Hence if the filtrate Group III or any of the succeeding groups are used to test for acid radicals, 
nitrates may fail to be shown and sulfites or sulfates may be reported although not present in the 
original unknown. 

(3) If arsenic is present, the filtrate from Group II metals should be acidified with about 5 cc. 
of concentrated hydrochloric acid and the solution heated to boiling and H 2 S passed through until 
no further ppt. is formed. This is to make certain that all arsenic has been removed and arsenic 
is best precipitated by H 2 S in strongly acid solutions. 

(4) When the solution is neutralized with NH 4 OH before testing for Group B acids, if a pre¬ 
cipitate is formed, filter and use the filtrate, as the precipitate contains only the hydroxides of the 
metals in the unknown. This filtrate may also be used in testing for the nitrate radical. 


196 


UNKNOWN REPORT IN ELEMENTARY QUALITATIVE ANALYSIS 


^ Report No, 


Sample No, 


Section No. 


Name.,. Date 

Sample (underline which, Solid, Liquid). 

Preliminary tests (color, odor, solubility, etc.). 


Tests for metals (Equations showing final proof of each). 


Tests for acids (Equations showing final proof of each). 


Probable combination of metal and acid. 


Summary of metals. 


Summary of acids. 


Accepted, 


Grade 


197 , 














































































% 


n 
























i 







UNKNOWN REPORT IN ELEMENTARY QUALITATIVE ANALYSIS 


Report No. 


Sample No, 


Section No. 


Name . Date 

Sample (underline which, Solid, Liquid). 

Preliminary tests (color, odor, solubility, etc.). 


Tests for metals (Equations showing final proof of each). 


Tests for acids (Equations showing final proof of each). 


Probable combination of metal and aeid. 


Summary of metals. 


Summary of acids. 


Accepted 


Grade 


199 

















































































































« 


* 
















« 




UNKNOWN REPORT IN ELEMENTARY QUALITATIVE ANALYSIS 


Report No. 


Sample No, 


Section No. 


Name . Date 

Sample (underline which, Solid, Liquid). 

Preliminary tests (color, odor, solubility, etc.). 


Tests for metals (Equations showing final proof of each). 


Tests for acids (Equations showing final proof of each). 


Probable combination of metal and acid. 


Summary of metals. 


Summary of acids. 


Accepted 


Grade 


201 











UNKNOWN REPORT IN ELEMENTARY QUALITATIVE ANALYSIS 


Report No, 


Sample No. 


Section No. 


Name. Date 

Sample (underline which, Solid, Liquid). 

Preliminary tests (color, odor, solubility, etc.). 


Tests for metals (Equations showing final proof of each). 


Tests for acids (Equations showing final proof of each). 


Probable combination of metal and acid. 


Summary of metals. 


Summary of acids. 


Accepted, 


Grade 





































. 









































































































UNKNOWN REPORT IN ELEMENTARY QUALITATIVE ANALYSIS 


Report No. 


Sample No, 


Section No, 


Name. Date 

Sample (underline which, Solid, Liquid). 

Preliminary tests (color, odor, solubility, etc.). 


Tests for metals (Equations showing final proof of each). 


Tests for acids (Equations showing final proof of each). 


Probable combination of metal and acid. 


Summary of metals. 


Summary of acids. 


Accepted, 


Grade 


205 











9 




■ 




































































































































♦ 


* 






















* 






UNKNOWN REPORT IN ELEMENTARY QUALITATIVE ANALYSIS 


Report No, 


Sample No, 


Section No. 


Name. Date 

Sample (underline which, Solid, Liquid). 

Preliminary tests (color, odor, solubility, etc.). 


Tests for metals (Equations showing final proof of each). 


Tests for acids (Equations showing final proof of each). 


Probable combination of metal and acid. 


Summary of metals. 


Summary of acids. 


Accepted, 


Grade 











UNKNOWN REPORT IN ELEMENTARY QUALITATIVE ANALYSIS 


Report No. 


Sample No. 


Section No, 


Name. Date 

Sample (underline which, Solid, Liquid). 

Preliminary tests (color, odor, solubility, etc.). 


Tests for metals (Equations showing final proof of each). 


Tests for acids (Equations showing final proof of each). 


Probable combination of metal and acid. 


Summary of metals. 


Summary of acids. 


Accepted 


Grade 
































































































UNKNOWN REPORT IN ELEMENTARY QUALITATIVE ANALYSIS 


Report No, 


Sample No, 


Section No. 


Name . Date 

Sample (underline which, Solid, Liquid). 

Preliminary tests (color, odor, solubility, etc.). 


Tests for metals (Equations showing final proof of each). 


Tests for acids (Equations showing final proof of each). 


Probable combination of metal and acid. 


Summary of metals. 


Summary of acids. 


Accepted, 


Grade 


211 












UNKNOWN REPORT IN ELEMENTARY QUALITATIVE ANALYSIS 


Report No. 


Sample No. 


Section No. 


Name . Date 

Sample (underline which, Solid, Liquid). 

Preliminary tests (color, odor, solubility, etc.). 


Tests for metals (Equations showing final proof of each). 


Tests for acids (Equations showing final proof of each). 


Probable combination of metal and acid. 


Summary of metals. 


Summary of acids. 


Accepted, 


Grade 


213 
















•« 





• • - • • • 








% 







- 




. 





v V 






* 


































































♦ 












































* <1 .1 ^ . * .* . 

















■) 




ft 



'■ 





* 




















* 


v ' 
























UNKNOWN REPORT IN ELEMENTARY QUALITATIVE ANALYSIS 


Report No, 


Sample No. 


Section No, 


Name. Date 

Sample (underline which, Solid, Liquid). 

Preliminary tests (color, odor, solubility, etc.). 


Tests for metals (Equations showing final proof of each). 


Tests for acids (Equations showing final proof of each). 


Probable combination of metal and acid. 


Summary of metals. 


Summary of acids. 


Accepted, 


Grade 
































APPENDIX 


SUGGESTED REVIEW OF WORK COVERED FIRST TERM IN RECITATION, 
LECTURE, AND LABORATORY 

(Be able to define and give illustrations) 


COMMON TERMS 


1 

Absolute density 

35 

Cathode 

2 

Absolute zero 

36 

Cation 

3 

Acid 

37 

Chemical property 

4 

Acid anhydride 

38 

Chemical reaction 

5 

Acid radical 

39 

Color 

6 

Acid salt 

40 

Combining weight 

7 

Active mass 

41 

Combustion 

8 

Affinity 

42 

Component 

9 

Air 

43 

Compound 

10 

Allotropic form 

44 

Constant boiling solution 

11 

Alkali 

45 

Critical pressure 

12 

Amorphous 

46 

Critical temperature 

13 

Analysis 

47 

Crystal 

14 

Anhydrous 

48 

Crystalline 

15 

Anion 

49 

Conductivity 

16 

Anode 

50 

Decomposition 

17 

Aqua regia 

51 

Decrepitation 

18 

Aqueous tension 

52 

Degree of dissociation 

19 

Atmosphere 

53 

Destructive distillation 

20 

Atom 

54 

Deliquescence 

21 

Base 

55 

Di-acid base 

22 

Basic anhydride 

56 

Dibasic acid 

23 

Basic radical 

57 

Diffusion 

24 

Basic salt 

58 

Dimorphous 

25 

Biological science 

59 

Direct union 

26 

Bivalent 

60 

Distillation 

27 

Boiling point 

61 

Double decomposition 

28 

British thermal unit 

62 

Ductility 

29 

By-product 

63 

Effervescence 

30 

Calorie 

64 

Efflorescence 

31 

Calorific value 

65 

Electrochemical series 

32 

Calorimeter 

66 

Electrode 

33 

Carbide 

67 

Electrolysis 

34 

Catalytic agent 

68 

Electrolyte 


217 


69 Electron 

70 Element 

71 Emulsion 

72 Endothermic reaction 

73 Energy- 

74 Equation 

75 Equilibrium 

76 Essential elements 

77 Evaporation 

78 Exothermic reaction 

79 Explosion 

80 Factors influencing reaction 

81 Filtrate 

82 Filtration 

83 Fluidity- 

84 Flux 

85 Formula 

86 Fractional distillation 

87 Freezing point 

88 Fusion 

89 Gas 

90 Gram atomic weight 

91 Gram molecular weight 

92 Hardness 

93 Heat 

94 Heat of condensation 

95 Heat of fusion 

96 Heat of vaporization 

97 Heterogeneous 

98 Homogeneous 

99 Hydrate 

100 Hydride 

101 Hydrocarbon 

102 Hydrolysis 

103 Hydroxylion 

104 Hydrolysis 

105 Indicator 

106 Inorganic chemistry 

107 Ion 

108 Ionization 

109 Isomorphous 

110 Kindling temperature 

111 Kinetic energy 

112 Law 

113 Liquid 

114 Luster 

115 Malleability 

116 Mass 

117 Matter 

118 Melting point 

119 Metal 


120 Metallic oxide 

121 Metalloid 

122 Metallurgy 

123 Mineral 

124 Mineral analysis 

125 Miscible 

126 Molar solution 

127 Molecule 

128 Mono-acid base 

129 Mono-basic acid 

130 Monoclinic form 

131 Nascent 

132 Neutralization 

133 Nitride 

134 Nitrogen cycle 

135 Nonessential elements 

136 Nonmetal 

137 Nonmetallic oxide 

138 Normality 

139 Normal salt 

140 Normal solution 

141 One atmosphere 

142 Opaque 

143 Ore 

144 Organic chemistry 

145 Osmotic pressure 

146 Oxidation 

147 Oxide 

148 Oxidizing agent 

149 Ozone 

150 Percentage strength 

151 Permanent hardness 

152 Peroxide 

153 Phlogiston theory 

154 Physical mixture 

155 Physical property 

156 Physical science 

157 Potability 

158 Potential energy 

159 Qualitative chemistry 

160 Quantitative chemistry 

161 Reducing agent 

162 Reduction 

163 Relative density 

164 Relative humidity 

165 Reversible reaction 

166 Rhombic form 

167 Salt 

168 Sanitary analysis 

169 Saturated solution 

170 Science 


218 


171 

172 

173 

174 

175 

17G 

177 

178 

179 

180 

181 

182 

183 

184 

185 

186 

187 

188 


Slag 

189 

Surface tension 

Smoke 

190 

Symbol 

Solid 

191 

Synthesis 

Solubility 

192 

Temperature 

Solute 

193 

Temporary hardness 

Solution 

194 

Tenacity 

Solvent 

195 

Theory 

Speed of oxidation 

196 

Titre 

Spontaneous combustion 

197 

Transformation of energy 

Standard temperature 

198 

Transparency 

Standard presssure 

199 

Univalent 

Strong electrolyte 

200 

Valence 

Structural formula 

201 

Vapor presssure 

Sublimation 

202 

Viscosity 

Substitution 

203 

Water of constitution 

Substance 

204 

Water of crystallization 

Supercooled solution 
Supersaturated solution 

205 

Weight 


219 




0 












¥ 


























# 






























* 



























# 




























r 
















































* 


t 



















GENERAL TYPE PROBLEMS 


DENSITY 

1. 2 cc. of a substance weighs 5 gms. Calculate its density. 

2. 5 gms. of a metal when placed in a graduated cylinder containing 10 cc. of water raised 
the water to 10.7 cc. Calculate the density of the metal. 

3. A block of metal 2 cm. by 3 cm. by 5 cm. weighs 90 gms. Calculate the density of the 
metal. 

4. If 4 cc. of aluminum weighs 10 gms. and 3 cc. of zinc weighs 22.5 gms., calculate the 
relative density of zinc using aluminum as the standard. 

5. Calculate the weight of 20 cc. of a substance having a density of 1.25 in the C. G. S. 
system. 

6. If 10 cc. of zinc weigh 70 gms., what is the density of zinc? 

7. If 91.2 gms. of iron displace 12 cc. of water, what is the density of iron? 

8. If the density of lead is 11.4, how much water would 68.4 gms. of lead displace? 

9. If the density of copper is 8.8, how much copper would be required to displace 100 cc. 

of water? 

10. 25 gms. of brass were put into 50 cc. of water, the water level was raised to 52.97 cc. 
mark. Find the density of the brass. 

11. (a) If density of lead is 11.4, what volume would 228.0 gms. occupy? 

(b) Find density of manganese if 25 cc. of it weighs 200.25 gms. 

12. (a) What is density of zinc if 100 cc. of it weighs 710 grams? 

(b) Find weight of 1 liter of oxygen if it has a density of 1.43. 

13. A lump of coal changed level of water in graduated cylinder from 12.5 cc. to 15 cc. mark. 
Find its volume and density if it weighed 6.25 gms. 

14. If 5 gms. of metal increases the volume of water in a graduate from 10 cc. to 12.5 cc., 
what is the density of the metal? 

15. If 10 gms. of a metal having a density of 1.5 is placed in 25 cc. of water, how much will 
the volume be increased? 

16. A piece of metal having the following dimensions 10 X 5 X 20 cms. weighs 1 kilogram. 
What is its density? 

17. A cylinder of metal having a diameter of 1 cm. and 5 cm. long weighs 15 gms. ^fyhat 
is its density? 

18. What is the mass of a piece of metal that has a density of 2.5 and increases the volume 
of water in a graduate 5 cc.? 


221 


MOLECULAR WEIGHT AND PERCENTAGE COMPOSITION 

Calculate molecular weights and percentage composition of the following: 

19. KClOs 

20. CuS0 4 .5H 2 0 

21. Na 2 B 4 0 7 

22. Cu(NH 3 ) 4 S0 4 

23. H 2 S0 4 

24. CaCOs 

25. Na 2 S 2 03 

26. Cu(N0 3 ) 2 

27. CaCl 2 

28. MgCOs 

29. A1 2 (S0 4 ) 3 

30. NH 4 C1 

31. CaHP0 4 

32. BaCl 2 

33. Na 2 C0 3 

34. CHsCOOH 

35. Ca 3 (P0 4 ) 2 

36. Ba0 2 

37. HgO 

38. Pb 3 0 4 

FORMULA FROM PERCENTAGE 

Calculate simplest formula of the following: 

39. Cl = 60.68%, Na = 39.31% 

40. As = 75.8%, O = 24.2% 

41. N = 82.35%, H = 17.65% 

42. Ca = 38.71%, P = 20%, O = 41.29% 

43. Na = 32.43%, S = 22.55%, O = 45.02% 

44. Pb = 86.6%, S = 13.4% 

45. C = 42.85%, O = 57.15% 

46. Pb = 68.3%, S = 10.55%, O = 21.15% 

47. Ca = 40%, C = 12%, O = 48% 

48. N = 28.17%, H = 8.11%, P = 20.82%, O = 42.9% 

49. C = 39.9%, H = 6.7%, O = 53.4% 

50. K = 26.58%, Cr = 35.39%, O = 38.03% 

51. H = 1.75%, S = 56.14%, O = 42.11% 

52. C = 73.8%, H = 8.7%, N = 17.1% 

53. C = 40.0%, O = 53.34%, H = 6.66& 


222 


Calculate true formula of the following: 


54. H = 5.88%, O = 94.12%, mol. wt. = 34 

55. C = 40.0%, O = 53.34%, H = 6.66%, mol. wt. = 90 

56. O = 23.8%, Cl = 52.5%, S = 23.7%, mol. wt. = 136.2 

57. O = 34.5%, H = 13.5%, C = 52.0%, mol. wt. = 46 

58. O = 74.0%, N = 26%, mol. wt. = 107.1 

59. H = 2.32%, C = 27.9%, N = 32.58%, O = 38.2%, mol. wt. = 129 

60. H = 2.22%, C = 26.66%, O = 71.11%, mol. wt. = 90 

61. N = 30.43%, O = 69.56%, mol. wt. = 92 

62. H = 5%, F = 95%, mol. wt. = 40 

63. H = 7%, C = 93%, mol. wt. = 26 

VOLUME AND WEIGHT 

64. Reduce to standard conditions: 

(1) 150 cc. at 780 mm. and 20° C. 

(2) 140 cc. at 745 mm. and 25° C. 

(3) 127 cc. at 720 mm. and 14° C. 

(4) 1000 cc. at 750 mm. and 20° C. 

(5) 1480 cc. at 765 mm. and 60° C. 

65. The temperature of a gas is 15° C. At what temperature would its volume be doubled? 

66. A gas measures 195 cc. at 740 mm. What is the pressure when the volume is 263 cc.? 

67. What volume of C0 2 at standard conditions can be prepared from 100 gms. of pure 
calcium carbonate? 

68. How many grams of ferrous sulfide are required to prepare 10 liters of H 2 S? (Stand¬ 
ard conditions.) 

69. How many grams of mercuric sulfide can be obtained by passing 100 cc. of H 2 S (20 C. 
and 755 mm. pressure) through a solution of mercuric nitrate? 

70. What volume of hydrogen can be prepared from 100 gms. of 95% zinc at 30 C. and 765 
mm.? 

71. What weight of KC10 3 is required to prepare 500 liters of oxygen at 30° C. and 15 
pounds pressure? 

72. What volume of hydrogen can be obtained by action of 6 gms. of zinc with dilute H 2 S0 4 ? 
Find weight of this volume obtained. 

73. Find weight of 1 liter of S0 2 ; 0 2 ; NII 3 . 

74. How much potassium chlorate will it take to prepare 20 liters 0 2 ? 

75. What weight of KC10 3 will be required to give 50 liters of oxygen under standard con¬ 
ditions? 

76. What weight of zinc and HC1 will be required to give 60 liters of hydrogen at 20° C. 
and 750 mm. pressure? 

77. What volume of oxygen at 15° C. and 780 mm. pressure can be obtained from heating 
100 gms. of HgO? 

78. What volume of S0 2 at 20° C. and 800 mm. pressure can be obtained by burning 50 gms. 
of sulfur? 


223 


79. If a given mass of a gas occupies 200 cc. at standard conditions, what volume will it 
occupy if collected over water at 20° C. and 75*.51 mm. pressure? 

80. If a given mass of a gas occupies 10 liters when collected over water at 27° C. and t 96.64 
mm. pressure, what volume will it occupy at standard conditions? 

81 If a given mass of a gas when measured over water at 20° C. and 767.51 mm. pressure 
was dried and placed in a cylinder at 27° C. and 3800 mm. pressure, how large was the tank if the 
original volume was 2000 liters? 

82. A cylinder filled with gas at 27° C. and 4560 mm. pressure was in a fire where the tem¬ 
perature rose to 1827° C., what pressure must the cylinder be able to withstand to prevent an 
explosion? 

83. Calculate the volume of gas liberated by the decomposition of 25 gms. of calcium car¬ 
bonate by 'heat. 

84. Calculate the weight of BaS0 4 that will be precipitated when 6.72 liters of S0 3 is run 
into an excess of barium chloride. 

85. Calculate the weight of mercuric sulfide that will be precipitated when 4.48 liters of 
hydrogen sulfide is passed into an excess of mercuric chloride. 

86. 450 cc. of gas measured at 22° C. and 735 mm. pressure will occupy what volume at 26° C. 
and 770 mm. pressure? 

87. A gas measured 4.3 liters over water at 20° C. and 757.4 mm. pressure. 'What will be 
its volume at standard conditions? 

88. A gas measured 350 cc. at standard conditions. What will be its volume at 50 C. and 
730 mm. pressure? 

89. What volume at 20° C. and 750 mm. will 100 liters at standard conditions be? 

90. What volume at standard conditions will be occupied by 50 liters of a gas at 30° C. 
and 800 mm. pressure? 

91. What volume at 50° C. and 720 mm. pressure will be occupied by 30 liters of a gas at 
20° C. and 850 mm. pressure? 

92. What volume at 30° C. and 900 mm. pressure will be occupied by 5000 cc. of a gas at 
20° C. and 750 mm.? 

93. What volume under standard conditions will be occupied by 60 cc. of a gas at - 10° 
and 3 atmospheres of pressure? 


COMBINING WEIGHTS 

94. If a compound has a formula of HgO, calculate the combining weight of Hg. 

95. If the atomic weight of Zn is 65.37 and its valence is 2, calculate its combining weight. 

96. If 10 gms. of calcium combines with 4 gms. of oxygen, calculate the combining weight of 

97. From the following data calculate the combining weights of the respective metals: 

(1) 0.5 gm. of calcium unites with 0.2 gm. of O. 

(2) 15 gms. of Hg unites with 1.2 gm. of O. 

(3) 2.3 gms. of Na liberates 0.1 gm. of H from H 2 0. 

(4) 0.03 gm. of Mg yields 30.4 cc. of H at 20° C. and 750 mm. 

98. Find combining weights of metals in following compounds: Fe 2 03 ; Cu 2 0; HgO; Pb0 2 . 

99. Calculate combining weight of N in NH 3 ; N0 2 ; N 2 0 2 ; and N 2 03 . 


224 


TITER AND NORMALITY 

100. If 10 cc. of H2SO4 having a Sp. Gr. of 1.25 requires 60 cc. of N/2 NaOH to neutralize it, 
calculate (1) its percentage strength, (2) normality, (3) titer. 

101. If 1 cc. of sulfuric acid, Sp. Gr. 1.8, when treated with BaCl 2 formed 4.2 gms. of BaS0 4 , 
calculate the percentage strength, normality, and titer of the acid. 

102. Calculate the volume of N/2 HC1 necessary to neutralize 100 cc. of N/5 KOH. 

103. If 50 cc. of 6N HC1 neutralizes 300 cc. of a solution of barium hydroxide, find normality 
and titer of base. 

104. A 10 gm. sample of H 2 S0 4 , Sp. Gr. 1.25 was neutralized with 40 cc. N/8 NaOH solution. 
Find titer, normality, and per cent of acid in sample. 

105. What vol. of con. HC1 solution (Sp. Gr. 1. 1; 37% HC1) wifi be required to neutralize 
1 liter of 5N NaOH solution? 

106. If 50 cc. of a N/10 NaOH neutralizes 35 cc. of an HC1 solution, what is the normality 
of the HC1 and the titer of each solution? 

107. If 35 cc. of a .25N Ca(OH) 2 is required to neutralize 75 cc. of HN0 3 solution, what is 
the normality of the HN0 3 and the titer of each solution? 

108. What volume of a N/4 H 2 S0 4 will be required to neutralize 125 cc. .05N Ba(OH) 2 
solution? 

109. What volume of a 2N KOH will be required to titrate 36 cc. of a N/2 H 2 S0 4 and what 
is the titer of each solution? 

110. What volume of a .37N H 3 P0 4 solution will be required to titrate 50 cc. of 2N NaOH 
solution and what is the titer of each solution? 

111. Suppose 37.5 cc. of a hydrochloric acid solution neutralizes 30 cc. of a sodium hydroxide 
solution, and that each cc. of sodium hydroxide solution contains 0.003 gm. of the base, what 
weight of hydrochloric acid is contained in the acid solution? 

112. What weight of sodium hydroxide is needed to neutralize 45 cc. of a sulfuric acid solu¬ 
tion having specific gravity of 1.3 and containing 32% (by weight) of pure H 2 S0 4 ? 

113. 10 cc. of N/10 HC1 (Sp. Gr. = 1.1) neutralized 15 cc. of KOH. Find normality, 
titer, and per cent of KOH in the sample. 

114. 10 gms. of vinegar was neutralized by 40 cc. of N/10 sodium hydroxide. Find per cent 
of acetic acid in the sample. 

115. 10 gm. of Ca(OH) 2 (Sp. Gr. = 1.2) was neutralized by 24 cc. of 3N H 3 P0 4 . Find the 
titer, normality, and per cent of Ca(OH) 2 in the sample. 


MISCELLANEOUS 

116. 2 gms. of H 2 S0 4 , Sp. Gr. 1.4, were neutralized by 14 cc. of normal NaOH solution. 
Calculate titer and normality of the acid. 

117. A 500 cc. bulb was filled over water with hydrogen at 20° C. and 740 mm. Calculate 
the weight of zinc used. Aq. ten. at 20 = 17.51 mm. 

118. 0.2 gm. of Na 2 C0 3 was neutralized by 15 cc. of N/10 HC1 solution. Calculate the 

purity of the Na 2 C0 3 . 

119. 75 cc. of water after removal of other salts was treated with NH 4 OH and Na 2 HP0 4 . 
The resulting MgNH 4 P0 4 was ignited and 0.647 gm. of Mg 2 P 2 0 7 obtaine . a c a e e per 
centage of MgS0 4 which was present in the sample of water. 

120. 100 gms. of limestone which is 93% pure is dissolved in HC1. What weight of salt is 

produced and what volume of gas at 30° C. and / SO mm.? , 

121. 2 gms. of 95% limestone is dissolved in H 2 S0 4 and the gas slowly bubbled through a 

225 


bulb containing KOH solution. The original weight of the bulb and KOH was 35.964 gms. 
What is the weight after evolution and absorption? 

122 . 55 pounds of metallic magnesium were produced by electrolysis of 800 pounds of fused 
carnallite. Calculate the purity of the carnallite. 

123. What weight of coal which tests 40% volatile matter and ash is required in the prepara¬ 
tion of 100 kgs. of Na 2 C0 3 by the LeBlanc process. What volume of HC1, sp. gr. 1.2, is produced 
in this process? 

124. (a) 0.86 gm. of a substance lowered the freezing point of 250 cc. of H 2 0 0.94 C. 
Calculate its molecular weight, (b) If volatile, what volume would this occupy under standard 
conditions? 

125. By the Victor Meyer method the following data was obtained: Wt. of sample = 0.56 
gms Initial reading in tube = 50 cc. Final reading in tube = 37 cc. Tempt, of H 2 0 = 20° C. 
Barometric reading = 755 mm. Aq. ten. at 20° = 17.51 mm. Calculate the molecular weight 
of the substance. 

126. What weight of IvC10 3 will be required for the preparation of oxygen sufficient to fill a 
tank holding 200 liters? 

127. In passing hydrogen over heated copper oxide 22.5 gms. of water was formed. W hat is 
the loss in the weight of the copper oxide? 

128. (a) What volume of oxygen measured at 20° C. and 740 mm. is evolved in the decompo¬ 
sition of 1 kg. of ordinary hydrogen peroxide? (b) What amount of heat is evolved in their de¬ 
composition? 

129. An automobile radiator contains 15 liters of w r ater. If 4 kgs. of grain alcohol is added, 
how much will the freezing point be lowered? 

130. 25 cc. of a normal solution of NaOH was required to neutralize the acid present in 5 cc. 
H 2 S0 4 . What is the strength of the acid in percentage? 

131. In an experiment by the freezing point method 2.2315 gms. of sugar lowered the fieezing 
point 0.122° C. when dissolved in 100 gms. of water. Calculate the molecular weight of the sugar. 

132. What volume of S0 2 will be generated by the action of HC1 on 20 gms. of sodium sulfite? 

133. If the atomic weights of Cu and Ag are 63 and 107, calculate their specific heats. 

134. 50 cc. N /8 Ca(OH ) 2 was required to neutralize 10 cc. FBPO 4 . Find titer of the base, 
normality of the acid, and percentage strength of acid. 

135. A certain limestone is 90% pure. What per cent of lime is present in this stone? Ca = 
40, C = 12, O = 16. 

136. Calculate the formula of the compound having the following composition: Alumina 
39.50%, silica 46.50%, water 13.98%. 

137. 76 liters of methane were collected over water at 27° C. and 776.65 mm. when aluminum 
carbide was treated with H 2 0. How much of this carbide, assuming it was 90% pure, was used? 
Aq. ten. = 26.65, A1 = 27, C = 12 . 

138. A plant makes 580 drums of acetylene per day. The capacity of each drum is 89.6 liters 
and the acetylene is put in at an average pressure of 20 atmospheres and an average temperat ure 
of 17° C. Assuming that the efficiency of operations is 80%, how many kilograms of limestone 
containing 84% CaC0 3 will be needed daily? 1 atmosphere = 760 mm., Ca = 40, C = 12 , O = 16. 

139. What weight of silver was precipitated as silver chloride when a solution of AgN0 3 was 
treated with 10 cc. of a N/3 solution of HC1? (Work to 4 decimals.) H = 1.008, Cl = 35.46, 
Ag = 107.88. 

140. How many kilograms of H 2 0 at 31° C. can be changed into steam at 100° C. when 1.080 
grams of A1 are burned to furnish the heat? Heat of combustion of A1 = 760.000 cal. Heat of 
vaporization of H 2 0 = 539 cal. per gram. A1 = 27. 

226 


141. 3.5 gms. of a substance was dissolved in 125 gms. of H 2 O. The freezing point of the 
solution was lowered 0.374° C. Calculate the molecular weight of the compound. 

142. How much sulfur must be used to make 48.480 gals, of calcium bisulfite solution 
whose Sp. Gr. is 1.25 and which contains 90% of the bisulfite? The liquor in this process is 
made by passing sulfur dioxide into a solution of limewater. Ca = 40, H = 1, S = 32, O = 16. 

143. 12.25 gms. of potassium chlorate is heated, (a) Calculate the weight of oxygen lib¬ 
erated. (b) Calculate volume of oxygen liberated at S. C. (c) Calculate volume of oxygen 
liberated if collected over water in laboratory at 20° C. and 757.51 mm. pressure. K = 39, 
Cl = 35.5, O = 16, aq. ten. at 20° = 17.51 mm. 

144. A 5 cc. sample of sulfuric acid having a Sp. Gr. of 1.2 required 100 cc. of N/5 NaOH to 
neutralize it. Calculate percentage, strength, normality, and titer of the acid. Na = 23, S = 32, 
O = 16 , H = L. 

145. If a 10-gram sample of a non-electrolyte lowers the freezing point of 50 cc. of water 
3.74° C., calculate the molecular weight of the substance. 

146. Calculate the M. W. of a substance from the following data of a Victor Meyer deter¬ 
mination: Wt. of liquid = 0.220 grams. Bar. pressure = 757.51. Vol. of air collected over 

water = .45 cc. Aq. ten. at 20° = 17.51. Tempt, of air = 20° C. 

147. Calculate the simplest formula of a compound having the following analysis: C = 
51.2%, O = 34.8%, Data: C = 12, O = 16, H = L. 

148. A compound contains 7.7% H and 92.3% C. In a Victor Meyer determination .1741 

gm. of the substance was used and 61.84 cc. of air were collected over water at 20° C. and 757.5 

mm. Calculate the true formula. C = 12, H = I, aq. ten. = 17.51 mm. 

149. A sulfuric acid plant has a daily output of 100 tons of acid containing 60% hydrogen 
sulfate. What weight of ferrous sulfide would be necessary if the ore is 95% pure and 5% of 
the S is lost in the process? Fe = 55.84, O = 16, H =-I, S = 32. 

150. Find per cent of acid in hydrobromic acid solution if 5 gms. of the acid were found to neu¬ 
tralize 10 cc. N/3 BaOH solution. 

151. Find molecular weight of the sugar if 2.231 gms. of the sugar dissolved in 100 cc. lowered 
freezing point of water .122° C. 

152. One gram zinc ore was dissolved and the zinc precipitated by hydrogen sulfide. Result¬ 
ing zinc sulfide weighed .41 gms. Calculate per cent of zinc in the ore. 

153. What volume of oxygen will be required to oxidize completely 25 liters of methane 
gas? Compare volume of C0 2 formed with volume of oxygen used. 

154. Find normality of NaOH solution that requires 10 cc. to neutralize 15 cc. H 2 S0 4 solution. 

155. A compound was found to have the following composition: K = 31.91%, Cl = 28.93%, 
O = 39.16%. Find simplest formula. What would formula be if mol. wt. was given as 245.12? 

156. What weight of sodium bicarbonate reacting with H 2 S0 4 will it require to prepare 100 1. 
of C0 2 measured at 15° C. and 737.8 mm. collected over water? Aq. ten. at 15° C. = 12.8 mm. 

157. If 20 cc. of acetylene gas weighs .0232 gms., find mol. weight of the gas. What weight 
of carbide would be required to furnish amount weighed? 

158. (a) Sp. heat of Fe = .112; combining weight = 27.92; find true atomic wt. (b) What 
would the density of iron be if 12 cc. weighed 93.6 gms.? 

159. If 1.42 gms. of a metal whose density is 7.1 is placed in a graduated cylinder contain¬ 
ing 5 cc. of water, what will be the new reading on the cylinder? 

160. A certain compound contains: Ca = 38.74%, P = 20.01%, and O = .4123%. Find 
the formula. 

161. How much limestone containing 85% CaC0 3 will be necessary to prepare 50 liters of 
CO2 measured at 745 mm. and 15° C.? 


227 


162. What weight of H 2 S0 4 will be necessary to precipitate the barium from a solution con¬ 
taining 20 gms. of the chloride? 

163. If 20 cc. of N/5 NaOH were required to neutralize a 15 cc. sample of H 3 P 4, 1 1 er > 

normality, and per cent of acid present. Density ot H 3 I 0 4 — 1.8. 

164. How much N/4 KOH would be required to neutralize 12 cc. of N/8 H 2 S0 4 ? 

165. 10 gals, of a certain water contains 12 gms. of calcium acid carbonate and 3 gms. of 
MgCU. What weights of Ca(OH) 2 and Na 2 C0 3 are necessary to soften 5000 gals, of this water. 

*166. 10 cc. of unknown H 2 S0 4 neutralizes 10.2 cc. of NaOH, titer .00492. Calculate titer 
and normality of the acid. 

167. What volume of ammonia at 20° C. and 750 mm. pressure can be obtained from 10 hters * 
of nitrogen at - 10° C. and 800 mm. pressure? 

168. What weight of Cu(N0 3 ) 2 can be prepared from 30 gms. of Cu and 100 gms. HN0 3 ? 

169. What weight of Fe 2 0 3 can be obtained on heating 50 gms. of iron in air? 

170. What weight of NaCl can be obtained on evaporating the solution made by 30 gms. of 
sodium reacting with an excess of HC1? 

171. What weight of nitrogen can be obtained, if 50 gms. of NaN0 2 is treated with an excess 
of NH 4 C1 and the yield is only 90% theoretical? 

172. If the yield is 75% theoretical, what weight of zinc and nitric acid will be required to 
give 100 gms. of Zn(N0 3 ) 2 ? 

173. If 2.84 gms. of a substance lowers the F. P. of 40 gms. of water 3.74° C., calculate its 
M. W. 

174. If .92 gms. of a substance raises the B. P. of 20 gms. of water .27° C., calculate its M. W. 

175 If 044 gms. of liquid when treated in a Victor Meyer apparatus displaced 24.62 cc. of 

air, if the experiment were run at 27° C. and 786.64 mm. pressure, calculate the M. W. of the 
substance in the gaseous state. 

176. If the Sp. Ht. of aluminum is .22, calculate its approximate atomic weight. 

177. If the Sp. Ht. of calcium is .1453, calculate its approximate atomic weight. 

178. What weight of limestone (80% pure CaC0 3 ) will it take to make 1 kgm. pure Ca(OH) 2 ? 

179. Find weight of rock salt required to produce 5 kgms. of pure NaOH if salt is 90% pure 
NaCl. 

180. How much air (21% O*) will it require to completely oxidize 250 gms. of carbon? 

181. Calculate the weight of sodium sulfate that can be prepared from 29.23 tons of sodium 
chloride if the yield is 90% of the theoretical. 

182. Calculate the weight of carbon dioxide that will be liberated when 5 gms. of calcium 
carbonate is treated with sulfuric acid. 

183. Calculate the weight of aluminum necessary to precipitate 12.714 gms. of copper. 

184. (1) What weight of sodium chloride is required to prepare 1 kg. of 40% hydrochloric acid? 

(2) What weight of 95% sulfuric acid would be required? 

185. How many grams of each product are formed when hydrochloric acid interacts with 
5 gms. of manganese dioxide? 

186. Ten tons of coke were burned and 35 tons of carbon dioxide were produced. Calculate 
the per cent of carbon in the coke. 

187. How many grams of calcium can be obtained from: 

(a) 170 gms. of calcium chloride? 

(b) 1 kg. of Iceland spar? 


228 


SOLUBILITIES OF VARIOUS SALTS 


It is evident that a knowledge of the solubility of compounds is an important part of our work 
in quahtative analysis. With this knowledge it is possible to predict the course of a reaction and 
to devise ways in which to prepare desired compounds. For precise information a dictionary of 
solubilities must be consulted, but it is possible to make a few general statements covering the 
most familiar classes of salts which will be of much assistance in understanding the reaction 
obtained by a qualitative analysis. These statements apply only to normal salts. Acid salts are 
apt to be more soluble than normal salts and basic salts less so. 

1. Nitrates. All nitrates are soluble. 

2. Chlorides. All chlorides are soluble except silver and mercurous chlorides. (Lead chloride 
is sparingly soluble.) 

3. Sulfates. All sulfates are soluble except those of barium, strontium, and lead. (Sulfates 
of silver and calcium are only moderately soluble.) 

4. Sulfides. All sulfides are insoluble except those of ammonium, sodium, and potassium. 
The sulfides of calcium, strontium, barium, and magnesium are insoluble in water, but are 
changed by hydrolysis into acid sulfides which are soluble. On this account they cannot be 
prepared by precipitation. 

5. Carbonates and Sulfites. All of these normal salts are insoluble except those of ammo¬ 
nium, sodium, and potassium. 

6. Acetates. All acetates are soluble. 


229 


TABLE OF SOLUBILITIES OF VARIOUS SALTS 



Chloride 

Bromide 

Iodide 

Sulfide 

Sulfate 

Sulfite 

Carbonate 

Phosphate 

Nitrate 

Acetate 

Pb 

1.49 

.60 

.08 

.004 

.004 

.Oil 

.Oil 

.0 4 1 

51.7 

50.0 

He 1 

.002 

. 0 5 4 

.041 


.006 




V.S. 

.75 

Ag 

•0 3 1 

,0 4 1 

. 0 6 3 

. 0 4 2 

.73 


.003 

. 0 3 6 

21.3 

1.04 

Hg 11 

6.9 

.23 

.005 

.002 

S 




V.S. 

30 

Bi 

Cl 

Q 


0 4 1 

s 

s 



s 

S 

b 

o 

V ft 


0 4 3 

20 0 




54.0 

8 

Cu 

Cd 

4o. U 

53.0 

V . 0 . 

49.0 

85.0 

.0»1 

76.0 




127.0 

V.S. 

Sn 

285.0 

s.s. 

.9 

.0 6 2 

19.0 



S.S. 


s.s. 

Sb 

601.6 

s 

S 

.0*1 

S 

s 



S 

s 

As 


Cl 


0 4 5 







S 

b 

no n 


0 3 8 

26 4 


.073 


45.0 


Fe 

A1 

74.39 
41.13 

oo . u 

s.s. 

S.S. 

S 

35.0 




V.S. 

s.s. 

Cr 

Ni 

i a n n 

200 0 



120.0 



s.s. 

s.s. 

s.s. 

JLbfcU. u 

38.0 

'56.0 

58.0 

. 0 3 3 

35.0 




48.0 

16.6 

Co 

32.0 

21.0 

64.0 

. 0 3 3 

35.0 




50.0 

S.S. 

Mn 

72.0 

59.0 

V.S. 

. 0 3 4 

62.0 


.013 


57.3 

3.0 

Zn 

204.0 

478.0 

419.0 

.0 3 6 

53.0 

.16 

.004 


118.0 

35. 0 

Ba 

37.2 

104.0 

201.0 

S 

. 0 3 2 

.01 

.002 

.01 

8.7 

69.0 

Sr 

51.1 

96.5 

169.0 

s 

.01 

.003 

.001 


66.2 

S.S. 

Ca 

73.2 

143.0 

200.0 

.2 

.2 

.004 

.001 

.01 

122.0 

34.0 

Mg 

55.8- 

103.0 

148.0 

S 

35.4 

1.2 

.1 

.02 

74.3 

S.S. 

Na 

35.9 

88.8 

178.0 

15.3 

16.8 

24.8 

19.4 

9.1 

84.0 

48.0 

K 

33.0 

66.0 

137.0 

S 

11.1 

100.0 

108.0 

249.0 

30.3 

245.0 

nh 4 

36.0 

41.0 

62.5 

V.S. 

75.0 

110.0 

50.0 

230.0 

185.0 

181.0 


Note. — The number in each square gives the number of grams of anhydrous salt held in 
solution by 100 cc. of watef. “S ” indicates soluble due to reaction in water solution, “V.S.” 
indicates very soluble, “S. S.” indicates slightly soluble. 


230 



























































EFFECT OF SPECIFIC REAGENTS 

(See Equations) 

HC1 Produces soluble chlorides with exception of Group I metals. Of these AgCl is soluble 

in NH 4 OH and PbCl 2 is soluble in hot water. 

H 2 S Produces insoluble sulfides of metals of Group I and II in slightly acid solution and of 

Group III and IV in slightly alkaline solution, except in cases of A1 and Cr where the 
hydroxides are formed. Some FeS may be precipitated in colloidal form in alkaline 
solution. The sulfides of metals of Groups V and VI decompose in solution forming 
soluble compounds. Sulfides of Fe, Mn, and Zn are soluble in dilute HC1. Sulfides of 
As, Sb, Sn are soluble in (NH 4 ) 2 Sx. All sulfides are soluble in 1:1 ELNO 3 except Hg, 
Ni, Co. These are soluble in aqua regia. 

HNO3 Produces soluble nitrates. In the case of Sn and Sb the insoluble metastannic acid or 
antimoniate may be produced and with As the soluble arsenious or arsenic acid may be 
formed. If a soluble sulfide or polysulfide is present, sulfur may be precipitated. 

H 2 S0 4 Produces soluble sulfates with the exception of Pb, Ba, Sr, and sparingly soluble 
Hg(ous) and Ca. 

NH 4 OH Produces soluble complex salts in cases of Ag, Cu, Cd, and metals of Groups IV and V. 

This is especially true in presence of excess NH 4 C1 in the solution. All other metals 
form hydroxides except Hg which gives an insoluble complex. 

NaOH Produces hydroxides of Pb, Zn, Al, As, Sn, and Cr and the oxide of Sb, all of which are 
soluble in excess of the alkali. Ag and Hg form insoluble oxides. Other metals, except 
those of the alkali family and dilute solutions of the alkaline earth family, form 
comparatively insoluble hydroxides. 

(NH 4 ) 2 C0 3 Produces insoluble carbonates, except in the case of Group VI metals. The carbon¬ 
ates of Ag, Cu, Cd, Co, Ni, and Zn redissolve in excess of the reagent. 


231 


TABLES OF WEIGHTS 

Metric System 


Milligram = 0.0154 grain 
Gram = 15.4323 grains 
Gram = 0.03527 ounce avoirdupois 
Gram — 0.0321 ounce troy 
Kilogram = 2.2046 pounds avoirdupois 
Kilogram = 2.6792 pounds troy 


Long ton = 
Short ton = 
Pound = 
Ounce = 
Grain = 


Pound = 
Ounce = 
Pennyweight = 
Grain = 
Ounce = 
Dram = 
Scruple = 


Avoirdupois 

2240 pounds = 1016.047 kilograms 
2000 pounds = 907.184 kilograms 
16 ounces = 7000 grains = 453.5924 grams 
437.5 grains = 2S.3495 grams 
64.798 milligrams = 0.06479 

Troy 

12 ounces = 5760 grains = 373.241 grams 
20 pennyweights = 480 grains = 31.103 grams 
24 grains = 1.555 grams 
64.7989 milligrams = 0.06479 gram 
8 drams = 480 grains = 31.1034 grams 
3 scruples = 60 grains = 3.8879 grams 
20 grains = 1.295 grams 


TABLES OF MEASURES 


Length 

Millimeter = 0.039 inch 
Centimeter = 0.393 inch 
Decimeter = 3.937 inches 
Meter = 39.37 inches 
Meter = 3.280 feet 
Meter = 1.0936 yards 
Inch = 2.540 centimeters 
Foot (12 inches) = 3.0480 decimeters 
Yard (3 feet) = 0.914 meter 
Mile (1760 yards) = 5280 feet 
Mile (1.609347 kilometers) = 1609.347 meters 


Surface 


Square millimeter 
Square centimeter 
Square decimeter 
Square decimeter 
Square meter 
Square meter (10.764 square feet) 


= 0.00155 square inch 
= 0.1549 square inch 
= 15.499 square inches 
= 0.1076 square foot 
= 1549.997 square inches 
= 1.195 square yards 


Volume 


Gallon (U.S.) 
Gallon (U.S.) 
Quart (U.S.) 
Pint (U.S.) 
Liter (U.S.) 


Cubic meter 
An imperial gallon (English) 


231 cubic inches 
3.785 liters 
0.946 liter 
0.473 liter 
2.113 pints (U.S.) 

1.0566 quarts (U.S.) 

0.264 gallon (U.S.) 

1.307 cubic yards 
35.314 cubic feet 
4.545 liters 

277.410 cubic inches (U.S.) 






* 









* 



























TABLE OF ACTIVITY OF METALS 


Potassium 

Sodium 

Barium 

Strontium 

Calcium 

Magnesium 


Aluminum 

Nickel 

Antimony 

Manganese 

Tin (Sn") 

Mercury (Hg") 

Zinc 

Lead 

Silver 

Cadmium 

Hydrogen 

Palladium 

Iron (Fe") 

Copper 

Platinum 

Cobalt 

Bismuth 

Gold 


DENSITIES AND MELTING POINTS OF SOME OF THE ELEMENTS 



Density 

Melting 

Point 

C° 

Aluminum. 

2.65 

658.7 

Antimony. 

6.70 

630. 

Arsenic. 

5.73 

850. 

Barium. 

3.75 

850. 

Bismuth... 

9.80 

271. 

Bromine. 

3.102 

-7.3 

Cadmium... 

8.64 

320.9 

Calcium. 

1.55 

810. 

Carbon, Diamond 

3.52 


Carbon, Graphite 

2.30 

3600. 

Chromium. 

7.30 

1615. 

Cobalt. 

8.60 

1480. 

Copper.. 

8.93 

1083. 

Gold. 

19.32 

1063. 

Iodine.. 

4.95 

113.5 

Iron.. 

7.86 

1530. 

Lead. 

11.37 

327.4 



Density 

Melting 

Point 

C° 

Magnesium. 

1.74 

651. 

Manganese. 

8.01 

1230. 

Mercury. 

13.56 

-38.87 

Nickel. 

8.90 

1452. 

Phosphorus, yellow 

1.83 

44. 

Platinum. 

21.50 

1755. 

Potassium.... 

0,862 

62.3 

Silicon.. 

2.35 

1420. 

Silver.. 

10.50 

960.5 

Sodium.. 

0.971 

97.5 

Strontium. 

2.54 

(?) 

Sulfur, monoclinic 

1.96 

119.2 

Sulfur, rhombic... 

2.06 

112.8 

Tin. 

7.30 

231.9 

Tungsten. 

18.72 

3400. 

Vanadium. 

6.02 

1720. 

Zinc.........- 

7.10 

419.4 































. ■ 




































